Concise Complex Analysis Solution Manual

Concise Complex AnalysisSolution of Exercise Problems Ai Shu Xue March 9, 2008 1 Contents 1 Calculus 3 2 Cauchy Integral Theorem and Cauchy Integral Formula 14 3 Theory of Series of Weierstrass 27 4 Riemann Mapping Theorem 52 5 Differential Geometry and Picard’s Theorem 53 6 A First Taste of Function Theory of Several Complex Variables 53 7 Elliptic Functions 53 8 The Riemann ζ-Function and The Prime Number Theory 53 2 This is a solution manual of selected exercise problems from Concise Complex Analysis, Second Edition, by Gong Sheng (World Scientific, 2007). This version solves the exercise problems in Chapter 1-3, except the following: Chapter 1 problem 37-42; Chapter 2 problem 47, 49; Chapter 3 problem 15 (xi). 1 Calculus 1. Proof. See, for example, Munkres [4], §38. 2. (1) Proof. |2i| = 2, arg(2i) = π 2 . |1 − i| = √ 2, arg(1 − i) = 7 4 π. |3 + 4i| = 5, arg(3 + 4i) = arctan 4 3 ≈ 0.9273 (Matlab command: atan(4/3)). | −5+12i| = 13, arg(−5+12i) = arccos _ − 5 13 _ ≈ 1.9656 (Matlab command: acos(−5/13)). (2) Proof. (1 + 3i) 3 = −26 −18i. 10 4−3i = 8 5 + 6 5 i. 2−3i 4+i = 5 17 − 14 17 i. (1 +i) n + (1 −i) n = 2 n 2 +1 cos n 4 π. (3) Proof. | −3i(2 −i)(3 + 2i)(1 +i)| = 3 √ 130. ¸ ¸ ¸ (4−3i)(2−i) (1+i)(1+3i) ¸ ¸ ¸ = 5· √ 5 √ 2· √ 10 = 5 2 . 3. Proof. Let θ = arctan 1 5 , α = arctan 1 239 . Then 5 − i = √ 26e −iθ , 1 + i = √ 2e i π 4 and (5 − i) 4 (1 + i) = 676 √ 2e i( π 4 −4θ) . Meanwhile (5 − i) 4 (1 + i) = 956 − 4i = 676 √ 2e −αi . So we must have π 4 − 4θ = −α + 2kπ, k ∈ Z, i.e. π 4 = 4 arctan 1 5 −arctan 1 239 +2kπ, k ∈ Z. Since 4 arctan 1 5 −arctan 1 239 +2π > 0− π 2 +2π = 7 4 π > π 4 and 4 arctan 1 5 − arctan 1 239 − 2π < 4 · π 2 − 2π = 0 < π 4 , we must have k = 0. Therefore π 4 = 4 arctan 1 5 − arctan 1 239 . 4. Proof. If z = x + yi, 1 ¯ z = x x 2 +y 2 + y x 2 +y 2 i. z 2 = x 2 − y 2 + 2xyi. 1+z 1−z = 1−x 2 −y 2 (1−x) 2 +y 2 + 2y (1−x) 2 +y 2 i. z z 2 +1 = x 3 +xy 2 +x+i(−y 3 −x 2 y+y) (x 2 −y 2 +1) 2 +4x 2 y 2 . 5. Proof. a = 1, b = α +iβ, c = γ +iδ. So ∆ = b 2 −4ac = (α 2 −β 2 −4γ) +i(2αβ −4δ) and z = −(α +iβ) ± √ ∆ 2 . 6. Proof. Denote arg z by θ, then z + r 2 z = re iθ + r 2 re iθ = r(e iθ + e −iθ ) = 2r cos θ = 2Rez, and z − r 2 z = re iθ − r 2 re iθ = r(e iθ −e −iθ ) = 2ir sinθ = 2iImz. 7. 3 Proof. If a = r 1 e iθ 1 and b = r 2 e iθ 2 , then ¸ ¸ ¸ ¸ a −b 1 − ¯ ab ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ r 1 −r 2 e i(θ 2 −θ 1 ) 1 −r 1 r 2 e i(θ 2 −θ 1 ) ¸ ¸ ¸ ¸ . Denote θ 2 −θ 1 by θ, we can reduce the problem to comparing |r 1 −r 2 e iθ | 2 and |1 −r 1 r 2 e iθ | 2 . Note |r 1 −r 2 e iθ | 2 = (r 1 −r 2 cos θ) 2 +r 2 2 sin 2 θ = r 2 1 −2r 1 r 2 cos θ +r 2 2 and |1 −r 1 r 2 e iθ | 2 = (1 −r 1 r 2 cos θ) 2 +r 2 1 r 2 2 sin 2 θ = 1 −2r 1 r 2 cos θ +r 2 1 r 2 2 . So |1 −r 1 r 2 e iθ | 2 −|r 1 −r 2 e iθ | 2 = (r 2 1 −1)(r 2 2 −1). This observation shows ¸ ¸ ¸ a−b 1−¯ ab ¸ ¸ ¸ = 1 if and only if at least one of a and b has modulus 1; ¸ ¸ ¸ a−b 1−¯ ab ¸ ¸ ¸ < 1 if |a| < 1 and |b| < 1. 8. Proof. We prove by induction. The equation clearly holds for n = 1. Assume it holds for all n with n ≤ N. Then ¸ ¸ ¸ ¸ ¸ N+1 i=1 a i b i ¸ ¸ ¸ ¸ ¸ 2 = ¸ ¸ ¸ ¸ ¸ N i=1 a i b i +a N+1 b N+1 ¸ ¸ ¸ ¸ ¸ 2 = _ N i=1 a i b i +a N+1 b N+1 __ N i=1 ¯ a i ¯ b i + ¯ a N+1 ¯ b N+1 _ = ¸ ¸ ¸ ¸ ¸ N i=1 a i b i ¸ ¸ ¸ ¸ ¸ 2 +|a N+1 | 2 |b N+1 | 2 + ¯ a N+1 ¯ b N+1 N i=1 a i b i +a N+1 b N+1 N i=1 ¯ a i ¯ b i = N i=1 |a i | 2 N i=1 |b i | 2 − i≤i<j≤N |a i ¯ b j −a j ¯ b i | 2 +|a N+1 | 2 |b N+1 | 2 +|b N+1 | 2 1≤i<j=N+1 |a i | 2 +|a N+1 | 2 1≤i<j=N+1 |b i | 2 − 1≤i<j=N+1 _ |a i | 2 |b N+1 | 2 +|a N+1 | 2 |b i | 2 −a i ¯ a N+1 b i ¯ b N+1 − ¯ a i a N+1 ¯ b i b N+1 ¸ = N+1 i=1 |a i | 2 N+1 i=1 |b i | 2 − 1≤i<j≤N |a i ¯ b j −a j ¯ b i | 2 − 1≤i<j=N+1 |a i ¯ b j −a j ¯ b i | 2 = N+1 i=1 |a i | 2 N+1 i=1 |b i | 2 − 1≤i<j≤N+1 |a i ¯ b j −a j ¯ b i | 2 . By method of mathematical induction, | n i=1 a i b i | 2 = n i=1 |a i | 2 n i=1 |b i | 2 − 1≤i<j≤n |a i ¯ b j −a j ¯ b i | 2 holds for all n ∈ N. Cauchy’s inequality follows directly from this equation. 9. Proof. We first consider the special case where a 1 = 0 and a 2 = 1. Then for a 3 = re iθ , a 1 , a 2 , a 3 are vertices of an equilateral triangle if and only if r = 1 and cos θ = 1 2 . Meanwhile the equation 1 + re 2iθ = re iθ is equivalent to _ 1 +r cos 2θ = r cos θ r sin2θ = r sinθ, which implies r = 1 and cos θ = 1 2 . So the equality is proven for the special case a 1 = 0, a 2 = 1. For general case, note a 2 1 + a 2 2 + a 2 3 = a 1 a 2 + a 2 a 3 + a 3 a 1 if and only if (a 2 − a 1 ) 2 + (a 3 − a 1 ) 2 = (a 2 −a 1 )(a 3 −a 1 ), which is further equivalent to 1 + _ a 3 −a 1 a 2 −a 1 _ 2 = a 3 −a 1 a 2 −a 1 . This shows the general case can be reduced to the special case of a 1 = 0 and a 2 = 1. Essentially, these reductions correspond to the composition of coordinate translation and coordinate rotation. 4 10. Proof. D := ¸ ¸ ¸ ¸ ¸ ¸ α ¯ α 1 β ¯ β 1 γ ¯ γ 1 ¸ ¸ ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ ¸ ¸ α −γ ¯ α − ¯ γ 1 β −γ ¯ β − ¯ γ 1 0 0 1 ¸ ¸ ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ α −γ ¯ α − ¯ γ β −γ ¯ β − ¯ γ ¸ ¸ ¸ ¸ . Suppose α −γ = r 1 e iθ 1 and β −γ = r 2 e iθ 2 . Then ¸ ¸ ¸ ¸ α −γ ¯ α − ¯ γ β −γ ¯ β − ¯ γ ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ r 1 e iθ 1 r 1 e −iθ 1 r 2 e iθ 2 r 2 e −iθ 2 ¸ ¸ ¸ ¸ = r 1 r 2 e i(θ 1 −θ 2 ) −r 1 r 2 e −i(θ 1 −θ 2 ) . So D = 0 if and only if θ 1 −θ 2 = kπ, which is equivalent to α, β, γ being colinear. Note the above reduction corresponds to the coordination transformation that places γ at origin. 11. Proof. We apply formula (1.11) (correction: the formula for x 2 should be x 2 = z−¯ z (1+|z| 2 )i ). If z = 1 −i, then x 1 = 2 3 , x 2 = − 2 3 , and x 3 = 2 3 . If z = 4 + 3i, then x 1 = 4 13 , x 2 = 3 13 , and x 3 = 12 13 . 12. Proof. Denote by x = (x 1 , x 2 , x 3 ) and y = (y 1 , y 2 , y 3 ) the points on the Riemann sphere S 2 corresponding to z 1 and z 2 , respectively. Then x and y are two ends points of a diagonal of S 2 if and only if x +y = 0. By formula (1.11), x +y = 0 is equivalent to _ ¸ ¸ _ ¸ ¸ _ z 1 +¯ z 1 1+|z 1 | 2 + z 2 +¯ z 2 1+|z 2 | 2 = 0 z 1 −¯ z 1 1+|z 1 | 2 + z 2 −¯ z 2 1+|z 2 | 2 = 0 |z 1 | 2 −1 |z 1 | 2 +1 + |z 2 | 2 −1 |z 2 | 2 +1 = 0. Solving the third equation gives |z 1 z 2 | = 1. So we can assume z 1 = re iα and z 2 = 1 r e iβ (r > 0). Then 1 +|z 1 | 2 = 1 +r 2 and 1 +|z 2 | 2 = 1 + 1 r 2 = 1 r 2 (1 +|z 1 | 2 ). So the other two equations become _ 2r cos α + 2r cos β = 0 2r sinα + 2r sinβ = 0, which is equivalent to _ cos α+β 2 cos α−β 2 = 0 sin α+β 2 cos α−β 2 = 0. So we must have 1 2 (α−β) = kπ + π 2 (k ∈ Z), i.e. α −β = 2kπ +π. So z 1 ¯ z 2 = e i(α−β) = e i(2kπ+π) = −1. 13. Proof. Suppose z ∈ C is on a circle. Then there exists z 0 ∈ C and R > 0, such that |z − z 0 | = R. Suppose z = x+iy and z 0 = x 0 +iy 0 . Then |z −z 0 | = R implies (x−x 0 ) 2 +(y −y 0 ) 2 = R 2 . After simplification, this can be written as |z| 2 + ¯ z 0 z +z 0 ¯ z +|z 0 | 2 −R 2 = 0. Let A = 1, B = ¯ z 0 and C = |z 0 | 2 −R 2 . Then A, C ∈ R and |B| 2 > AC. Conversely, if A|z| 2 +Bz + ¯ B¯ z +C = 0 is the equation satisfied by z, we have two cases to consider: (i) A = 0; (ii) A = 0. In case (i), A|z| 2 + Bz + ¯ B¯ z + C = 0 can be written as |z| 2 + B A z + ¯ B A ¯ z + |B| 2 A 2 = |B| 2 A 2 − C A > 0. Define z 0 = ¯ B A and R = _ |B| 2 A 2 − C A . The equation can be written as |z| 2 +¯ z 0 z+z 0 ¯ z+|z 0 | 2 = R 2 , which is equivalent to |z −z 0 | = R. So z is on a circle. 5 In case (ii), the equation Bz + ¯ B¯ z + C = 0 stands for a straight line in C. Indeed, suppose B = a + ib and z = x + iy, then Bz + ¯ B¯ z + C = 0 becomes 2ax − 2by + C = 0. If we regard straight lines as circles, A|z| 2 +Bz + ¯ B ¯ +C = 0 represents a circle in case (ii) as well. To see the necessary and sufficient condition for the image of this circle on the Riemann sphere S 2 to be a great circle, we suppose a generic point z on this circle corresponds to a point (x 1 , x 2 , x 3 ) on the Riemann sphere S 2 . Then by formula (1.10) z = x 1 +ix 2 1−x 3 , we can write the equation A|z| 2 + Bz + ¯ B¯ z + C = 0 as (B+ ¯ B)x 1 +i( ¯ B−B)x 2 +(A−C)x 3 = −(A+C). (x 1 , x 2 , x 3 ) falls on a great circle if and only if (x 1 , x 2 , x 3 ) lies on the intersection of S 2 and a plane that contains (0, 0, 0), which is equivalent to A + C = 0 by the equation (B + ¯ B)x 1 +i( ¯ B −B)x 2 + (A−C)x 3 = −(A+C). 14. Proof. Suppose Z 1 = (x 1 , x 2 , x 3 ) and Z 2 = (y 1 , y 2 , y 3 ). Then by formula (1.11), d(z 1 , z 2 ) 2 = (x 1 −y 1 ) 2 + (x 2 −y 2 ) 2 + (x 3 −y 3 ) 2 = 2 −2(x 1 y 1 +x 2 y 2 +x 3 y 3 ) = 2 −2 _ z 1 + ¯ z 1 1 +|z 1 | 2 · z 2 + ¯ z 2 1 +|z 2 | 2 + z 1 − ¯ z 1 i(1 +|z 1 | 2 ) · z 2 − ¯ z 2 i(1 +|z 2 | 2 ) + |z 1 | 2 −1 |z 1 | 2 + 1 · |z 2 | 2 −1 |z 2 | 2 + 1 _ = 4 |z 1 | 2 +|z 2 | 2 −z 1 ¯ z 2 − ¯ z 1 z 2 (1 +|z 1 | 2 )(1 +|z 2 | 2 ) = 4 (z 1 −z 2 )(¯ z 1 − ¯ z 2 ) (1 +|z 1 | 2 )(1 +|z 2 | 2 ) . So d(z 1 , z 2 ) = 2|z 1 −z 2 | √ (1+|z 1 | 2 )(1+|z 2 | 2 ) . Consequently, d(z 1 , ∞) = 2 lim z 2 →∞ ¸ ¸ ¸ z 1 z 2 −1 ¸ ¸ ¸ _ (1 +|z 1 | 2 ) _ 1 |z 2 | 2 + 1 _ = 2 _ 1 +|z 1 | 2 . 15. Proof. (i) The distance between z and z 1 is no greater than the distance between z and z 2 . (ii) z −z 1 and z −z 2 are perpendicular to each other. (iii) The angle between z +i and z −i is less than π 4 . (iv) z is on an ellipse with c and −c as the foci. 16. Proof. Regard C as R 2 and apply Heine-Borel theorem and Bolzano-Weierstrass theorem for R n (n ∈ N). 17. Proof. Note max{|Rez n −Rez 0 |, |Imz n −Imz 0 |} ≤ |z −z 0 | ≤ |Rez n −Rez 0 | +|Imz n −Imz 0 |. 18. (1) 6 Proof. We suppose |a|, |b| < ∞. Then (z n ) n and (z n ) n are bounded sequences. Denote by M a common bound for both sequences. ∀ε > 0, ∃N, so that for any n > N, max{|z n −a|, |z n −b|} < ε M . Then for any n satisfying n > N _ 1 + M 2 2 _ (define z 0 = z 0 = 0) ¸ ¸ ¸ ¸ ¸ 1 n n k=1 z k z n−k −ab ¸ ¸ ¸ ¸ ¸ < ¸ ¸ ¸ ¸ ¸ 1 n N k=1 z k z n−k ¸ ¸ ¸ ¸ ¸ + ¸ ¸ ¸ ¸ ¸ 1 n n k=N+1 z k z n−k −ab ¸ ¸ ¸ ¸ ¸ < NM 2 n + ¸ ¸ ¸ ¸ ¸ 1 n n k=N+1 (z k −a)z n−k ¸ ¸ ¸ ¸ ¸ + ¸ ¸ ¸ ¸ ¸ a n n k=N+1 z n−k −ab ¸ ¸ ¸ ¸ ¸ < ε +ε +|a| ¸ ¸ ¸ ¸ ¸ 1 n n−N−1 k=0 z k −b ¸ ¸ ¸ ¸ ¸ = 2ε +|a| ¸ ¸ ¸ ¸ ¸ n −N −1 n · 1 n −N −1 n−N−1 k=0 z k −b ¸ ¸ ¸ ¸ ¸ . Taking upper limits on both sides, we get lim n→∞ ¸ ¸ ¸ ¸ ¸ 1 n n k=1 z k z n−k −ab ¸ ¸ ¸ ¸ ¸ ≤ 2ε + 0 = 2ε. Since ε is arbitrary, we conclude lim n→∞ ¸ ¸ 1 n n k=1 z k z n−k −ab ¸ ¸ = 0, i.e. lim n→∞ 1 n n k=1 z k z n−k = ab. (2) Proof. Apply the result of (1) with z n ≡ 1. 19. Proof. (i) f(z) = √ z¯ z is a function of both z and ¯ z. So f is not differentiable. (ii) f(z) = ¯ z is a function of ¯ z, so it is not differentiable. (iii) f(z) = Rez = 1 2 (z + ¯ z) is a function of both z and ¯ z. So f is not differentiable. 20. Proof. By chain rule, ∂ ∂¯ z g(f(z)) = g (f(z)) ∂f(z) ∂¯ z = 0. So g(f(z)) is also holomorphic. 21. (1) Proof. Let u(x, y) = Ref(z) and v(x, y) = Imf(z). Then by C-R equations f (z) = u x (x, y) + iv x (x, y) = v y (x, y) − iu y (x, y). So f (z) ≡ 0 if and only if v x = v y = u x = u y ≡ 0. By results for functions of real variables, we conclude u and v are constants on D. so f is a constant function on D. (2) Proof. (i) and (ii) are direct corollaries of C-R equations. For (iii), we assume without loss of generality that f(z) ≡ 0 on D. We note |f(z)| 2 = u 2 (z) + v 2 (z). So by the C-R equations, 0 = ∂ ∂x |f(z)| 2 = 2uu x + 2vv x = 2uu x − 2vu y and 0 = ∂ ∂y |f(z)| 2 = 2uu y + 2vv y = 2uu y + 2vu x , i.e. _ u −v v u _ _ u x u y _ = 0. Since f ≡ 0, _ u −v v u _ is invertible on D 1 := {z ∈ D : f(z) = 0}. So u x = u y ≡ 0 on D 1 . By the C-R equations v x = v y ≡ 0 on D 1 . So f is a constant function on each simply 7 connected component of D 1 . Since D 2 := {z ∈ D : f(z) = 0} has no accumulation points in D (see Theorem 2.13), f must be identical to the same constant throughout D. For (iv), we assume arg f(z) ≡ θ. Then g(z) := e −iθ f(z) is also holomorphic and f is a constant function if and only if g is a constant function. So without loss of generality, we can assume arg f(z) ≡ 0. Then Imf(z) ≡ 0. By C-R equations, Ref(z) is a constant too. Combined, we can conclude f is identically equal to a constant on D. 22. Proof. The tangent vector a(x, y) of the curve u(x, y) = c 1 at point (x, y) is perpendicular to (u x , u y ); the tangent vector b(x, y) of the curve v(x, y) = c 2 at point (x, y) is perpendicular to (v x , v y ). Since (v y , −v x ) ⊥ (v x , v y ) and (u x , u y ) = (v y , −v x ) by C-R equations, we must have a(x, y) ⊥ b(x, y), which means the curves u(x, y) = c 1 and v(x, y) = c 2 are orthogonal. 23. (1) Proof. Since _ x = r cos θ y = r sinθ , we have _ ∂ ∂r ∂ ∂θ _ = _ cos θ sinθ −r sinθ r cos θ _ _ ∂ ∂x ∂ ∂y _ := A(θ, r) _ ∂ ∂x ∂ ∂y _ . It’s easy to see A −1 (θ, r) = 1 r _ r cos θ −sinθ r sinθ cos θ _ . Writing the Cauchy-Riemann equations in matrix form, we get _ ∂ ∂x ∂ ∂y _ u = _ 0 1 −1 0 _ _ ∂ ∂x ∂ ∂y _ v. Therefore, under the polar coordinate, the Cauchy-Riemann equations become _ ∂ ∂r ∂ ∂θ _ u = A(θ, r) _ ∂ ∂x ∂ ∂y _ u = A(θ, r) _ 0 1 −1 0 _ _ ∂ ∂x ∂ ∂y _ v = A(θ, r) _ 0 1 −1 0 _ A −1 (θ, r) _ ∂ ∂r ∂ ∂θ _ v = _ 0 1 r −r 0 _ _ ∂ ∂r ∂ ∂θ _ v. Fix z = re iθ , then f (z) = lim ∆r→0 u(r + ∆r, θ) +iv(r + ∆r, θ) −[u(r, θ) +iv(r, θ)] ∆r · e iθ = ∂u ∂r e −iθ +i ∂v ∂v e −iθ = r z _ ∂u ∂r +i ∂v ∂r _ . (2) Proof. If f(z) = R(cos φ +i sinφ), then u = Rcos φ and v = Rsinφ. So u r = ∂R ∂r cos φ−Rsinφ ∂φ ∂r , v r = ∂R ∂r sinφ+Rcos φ ∂φ ∂r , u θ = ∂R ∂θ cos φ−Rsinφ ∂φ ∂θ , v θ = ∂R ∂θ sinφ+Rcos φ ∂φ ∂θ . Writing the above equalities in matrix form, we have _ u r v r _ = A _ ∂R ∂r ∂φ ∂r _ , _ u θ v θ _ = A _ ∂R ∂θ ∂φ ∂θ _ , where A = _ cos φ −Rsinφ sinφ Rcos φ _ . Therefore the C-R equations become A _ ∂R ∂r ∂φ ∂r _ = _ u r v r _ = _ 0 1 r − 1 r 0 _ _ u θ v θ _ = _ 0 1 r − 1 r 0 _ A _ ∂R ∂θ ∂φ ∂θ _ . 8 Note A −1 = 1 R _ Rcos φ Rsinφ sinφ Rcos φ _ . So A −1 _ 0 1 r − 1 r 0 _ A = _ 0 R r − 1 Rr 0 _ . Therefore _ ∂R ∂r ∂φ ∂r _ = _ 0 R r − 1 Rr 0 _ _ ∂R ∂θ ∂φ ∂θ _ , i.e. _ ∂R ∂r = R r ∂φ ∂θ ∂R ∂θ = −Rr ∂φ ∂r . 24. Proof. By the C-R equations, J = ¸ ¸ ¸ ¸ u x u y v x v y ¸ ¸ ¸ ¸ = u x v y −v x u y = u x · u x −v x · (−v x ) = u 2 x +v 2 x = |f (z)| 2 . J is the area of f(D), see, for example, Munkres [4], §21 The volume of a parallelepiped. 25. (1) Proof. We can parameterize γ with γ(t) = t +ti (0 ≤ t ≤ 1). So _ γ xdz = _ 1 0 t(dt +idt) = 1 2 + 1 2 i. (2) Proof. _ γ |z −1||dz| = _ 2π 0 |e it −1|dt = _ 2π 0 √ 2 −2 cos tdt = √ 2 _ 2π 0 ¸ ¸ ¸ ¸ sin _ π 4 − t 2 _ −cos _ π 4 − t 2 _¸ ¸ ¸ ¸ dt = √ 2 _ π 4 − 3 4 π |sinθ −cos θ| dθ = √ 2 _ π 4 − 3 4 π (cos θ −sinθ) dθ4. (3) Proof. _ γ dz z −a = _ 2π 0 iRe it Re it dt = 2πi. 26. Proof. We first recall sinh z = e z −e −z 2 and coshz = e z +e −z 2 . Then straightforward calculations show cos(x +iy) = coshy cos x −i sinhy sinx, sin(x +iy) = coshy sinx +i sinhy cos x. By this result, we have siniz = e i(iz) −e −i(iz) 2i = e −z −e z 2i = sinhz, cos iz = e i(iz) +e −i(iz) 2 = e −z +e z 2 = cosh z, (sin z) = _ e iz −e −iz 2i _ = e iz +e −iz 2 = cos z. 9 and cos z 1 cos z 2 −sinz 1 sinz 2 = e iz 1 +e −iz 1 2 e iz 2 +e −iz 2 2 − e iz 1 −e −iz 1 2i e iz 2 −e −iz 2 2i = e i(z 1 +z 2 ) +e −i(z 1 +z 2 ) +e i(z 1 −z 2 ) +e −i(z 1 −z 2 ) 4 − e i(z 1 +z 2 ) +e −i(z 1 +z 2 ) −e i(z 1 −z 2 ) −e −i(z 1 −z 2 ) −4 = 2e i(z 1 +z 2 ) + 2e −i(z 1 +z 2 ) 4 = cos(z 1 +z 2 ). 27. Proof. sini = sinh1. cos(2 +i) = cosh1 cos 2 −i sinh1 sin2. tan(1 +i) = sin 1 cos 1+i sinh 1 cosh 1 cos 2 cosh 2 1 cos 2 1+sinh 2 1 sin 2 1 . 2 i = e −2kπ [cos(log 2) +i sin(log 2)], k ∈ Z. i i = e − π 2 +2kπ , k ∈ Z. (−1) 2i = e −2(2k+1)π , k ∈ Z. log(2 −3i) = log √ 13 +i[2kπ −arctan 3 2 ], k ∈ Z. arccos 1 4 (3 +i) = 3+i± √ 10e i( π 2 − 1 2 arctan 3 4 ) 4 . 28. Proof. e π 2 i = i, e − 2π 3 i = 1 2 − √ 3 2 i. log i = (2kπ + π 2 )i, k ∈ Z. 29. Proof. z z = e z log z = e (x+iy) log(x+iy) = e x 2 log(x 2 +y 2 )−y(2kπ+arctan y x ) ·e i[ y 2 log(x 2 +y 2 )+x(2kπ+arctan y x )] , k ∈ Z. 30. Proof. The root of z n = a are |a| 1 n e i 2kπ+arg a n , k = 0, 1, · · · , n −1. 31. (1) Proof. If u n (z) = Ref n (z) and v n (z) = Imf n (z), then by max{|u n (z) −u m (z)|, |v n (z) −v m (z)|} ≤ |f n (z) − f m (z)| ≤ |u n (z) −u m (z)| +|v n (z) −v m (z)|, the Cauchy criterion for convergence for complex field is reduced to that for real numbers. (2) Proof. Note | m n=k f n (z)| ≤ M m n=k a n . So ∞ n=1 f n (z) converges uniformly if ∞ n=1 a n converges. 32. (i) Proof. Since lim n→∞ n 1 n 2 = e lim n→∞ ln n n 2 = 1, Hadamard’s formula implies R = 1. (ii) Proof. Since lim n→∞ n ln n n = e lim n→∞ [ ln n ·ln n] 1, Hadamard’s formula implies R = 1. (iii) 10 Proof. By Stirling’s formula n! ∼ √ 2πn _ n e _ n , we have lim n→∞ _ n! n n _ 1 n = lim n→∞ e 1 n ln n! n n = lim n→∞ e 1 n [ln(n!)−ln √ 2πn( n e ) n ] · lim n→∞ e 1 n ln √ 2πn( n e ) n −ln n = e −1 . So Hadamard’s formula implies R = e. (iv) Proof. Since lim n→∞ n = ∞, Hadamard’s formula implies R = 0. 33. Proof. Let u(z) = Ref(z) and v(z) = Imf(z). Then f is uniformly convergent on a set A if and only if u and v are uniformly convergent on A. Since dz = dx+idy, we can reduce the theorem to that of real-valued functions of real variables. 34. (1) Proof. [f(z)e −z ] = f (z)e −z − f(z)e −z = 0. So f(z) = Ce z for some constant C. f(0) = 1 dictates C = 1. (2) Proof. f (z) = lim ω→0 f(z+ω)−f(z) ω = lim ω→0 f(z) f(ω)−1 ω = f(z)f (0) = f(z). By part (1), we conclude f(z) = e z . 35. (1) Proof. [e f(z) ] = f (z)e f(z) = 1. So e f(z) = 1. (2) Proof. It’s clear f(1) = 0. So f (z) = lim ω→0 f(z +zω) −f(z) zω = lim ω→0 f(z(1 +ω)) −f(z) zω = 1 z lim ω→0 f(1 +ω) ω = 1 z f (1) = 1 z . Therefore _ 1 z e f(z) _ = − 1 z 2 e f(z) + 1 z e f(z) f (z) = 0. This shows e f(z) = z. 36. Proof. 2[ϕ(z 1 ) − ϕ(z 2 )] = (z 1 − z 2 ) _ 1 − 1 z 1 z 2 _ . So ϕ(z) is univalent in any region that makes z 1 z 2 = 1 whenever z 1 = z 2 . Since z 1 z 2 = 1 implies |z 1 | = 1 |z 2 | and arg z 1 = −arg z 2 , we conclude ϕ is univalent in any of the four regions in the problem. 43. Proof. This is straightforward from Problem 24. 44. 11 Proof. We first use Abel’s Lemma on partial sum: for n > m, n k=m+1 a k b k = n k=m+1 (S k −S k−1 )b k = S b b n−1 + n k=m+1 S k (b k −b k+1 ) −S m b m+1 . Denote by K a bound of the sequence (S n ) ∞ n=1 , then |S n b n+1 | ≤ K|b n+1 |, |S m b m+1 | ≤ K|b m+1 |, and | n k=m+1 S k (b k − b k+1 )| ≤ K n k=m+1 |b k − b k+1 |. Since lim n→∞ b n = 0 and ∞ n=1 |b n − b n+1 | < ∞, we conclude n k=m+1 a k b k can be arbitrarily small if m is sufficiently large. So under conditions (1)-(3), ∞ n=1 a k b k is convergent. In the Dirichlet criterion, we require (1) (S n ) ∞ n=1 is bounded; (2) lim n→∞ b n = 0; (3’) (b n ) ∞ n=1 is monotone. Clearly, (2)+(3’) imply (3), so the Dirichlet criterion is a special case of the result in current problem. In the Abel criterion, we require (1”) (S n ) ∞ n=1 is convergent; (2”) (b n ) ∞ n=1 is bounded; (3”) (b n ) ∞ n=1 is monotone. Define b n = b n − b, where b = lim n→∞ b n is a finite number. Then (S n ) ∞ n=1 satisfies (1) and (b n ) ∞ n=1 satisfies (2) and (3’). By the Dirichlet criterion, ∞ n=1 a n b n converges. Note N n=1 a n b n = N n=1 a n b n +bS N . By condition (1”), we conclude ∞ n=1 a n b n converges. Remark 1. The result in this problem is the so-called Abel-Dedekind-Dirichlet Theorem. 45. (1) Proof. For any ε ∈ (0, 1 −q), there exists N ∈ N, such that for any n ≥ N, |a n | 1 n < q +ε < 1. By comparing ∞ n=1 |a n | and ∞ n=1 (q +ε) n , we conclude ∞ n=1 a n is absolutely convergent. (2) Proof. By definition of upper limit and the fact q > 1, we can find ε > 0 and infinitely many n, such that |a n | 1 n > q −ε > 1. Therefore lim n→∞ |a n | = ∞ and ∞ n=1 a n is divergent. 46. Proof. If q < 1, for any ε ∈ (0, 1 −q), we can find N ∈ N, such that for any n ≥ N, |a n+1 | |a n | < q + ε < 1. So |a N+k | ≤ |a N |(q +ε) k , ∀k ≥ 0. Since ∞ k=1 (q +ε) k is convergent, we conclude ∞ n=1 a n is also convergent. If q > 1, we have two cases to consider. In the first case, the upper limit is assumed to be a limit. Then we can find ε > 0 such that q − ε > 1 and for n sufficiently large, |a n+1 | > |a n |(q − ε) > |a n |. This implies lim n→∞ a n = 0, so the series is divergent. In the second case, the upper limit is assumed not to be a limit. Then we can manufacture counter examples where the series is convergent. Indeed, consider (k ≥ 0) a n = _ 1 (k+1) 2 , if n = 2k + 1 2 (k+1) 2 , if n = 2k + 2. Then ∞ n=1 a n is convergent and lim n→∞ ¸ ¸ ¸ a n+1 a n ¸ ¸ ¸ = 2. 47. 12 Proof. We choose ε > 0, such that lim n→∞ n _ | a n+1 a n | −1 _ < −1 −2ε. By the definition of upper limit, there exists N 1 ∈ N, such that for any n ≥ N 1 , n _ | a n+1 a n | −1 _ < −1 −2ε, i.e. |a n+1 | |a n | < 1 − 1+2ε n . Define b n = 1 n 1+ε . Then b n+1 b n = n ε (n + 1) 1+ε = _ 1 − 1 n + 1 _ 1+ε = 1 − 1 +ε n + 1 +O _ 1 (n + 1) 2 _ . So there exists N 2 ∈ N, such that for n ≥ N 2 , b n+1 b n > 1 − 1+2ε n+1 . Define N = max{N 1 , N 2 }, then for any n ≥ N, a n+1 a n < b n+1 b n . In particular, we have ¸ ¸ ¸ ¸ a N+1 a N ¸ ¸ ¸ ¸ < b N+1 b N , ¸ ¸ ¸ ¸ a N+2 a N+1 ¸ ¸ ¸ ¸ < b N+2 b N+1 , · · · , ¸ ¸ ¸ ¸ a N+k a N+k−1 ¸ ¸ ¸ ¸ < b N+k b N+k−1 . Multiply these inequalities, we get |a N+k | ≤ |a N | b N b N+k . Since ∞ n=1 b n is convergent, we conclude ∞ n=1 a n must also converge. 1 48. (1) Proof. Since (a n ) ∞ n=1 monotonically decreases to 0, for n sufficiently large, lna n < 0 and lim n→∞ ln a n n ≤ 0. So lim n→∞ a 1 n n = e lim n→∞ ln a n n ≤ e 0 = 1. By Hadamard’s formula, R ≥ 1. (2) Proof. The claims seems problematic. For a counter example, assume R > 1 and a n = 1 R n . If z = Re iθ with θ = 0, we have N n=0 a n z n = N n=0 1 R n · R n e inθ = N n=0 cos nθ +i N n=0 sinnθ = sin (N+1)θ 2 cos Nθ 2 sin θ 2 +i sin (N+1)θ 2 sin Nθ 2 sin θ 2 . So ∞ n=0 a n z n is not convergent. 49. Proof. By definition of uniform convergence, for any ε > 0, there exists N ∈ N, such that for any m ≥ N, ∞ n=m |a n ||z −z 0 | n < ε, ∀z ∈ D. In particular, for any z ∗ ∈ ∂D, by letting z → z ∗ , we get ∞ n=m |a n ||z ∗ − z 0 | n ≤ ε. That is, for this given ε, we can pick up N, such that for any m ≥ N, ∞ n=m |a n ||z − z 0 | n ≤ ε, ∀z ∈ ¯ D. This is exactly the definition of uniform convergence on ¯ D. 50. Proof. We choose a sequence (r n ) ∞ n=1 such that r n ∈ (0, 1) and lim n→∞ r n = 1. Define β n = n k=0 a k z k 0 − f(r n z 0 ). We show lim n→∞ β n = 0. Indeed, we note |β n | = ¸ ¸ ¸ ¸ ¸ n k=0 a k z k 0 − ∞ k=0 a k (r n z 0 ) k ¸ ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ ¸ n k=0 a k (1 −r k n )z k 0 − ∞ k=n+1 a k (r n z 0 ) k ¸ ¸ ¸ ¸ ¸ ≤ n k=0 |a k |(1 −r n )(1 +r n +· · · +r k−1 n ) + ∞ k=n+1 k n |a k |r k n ≤ n k=0 |a k |(1 −r n )k + ε n+1 n 1 1 −r n , 1 It seems the condition lim n→∞ a n+1 a n = 1 is redundant. 13 where ε n = k≥n k|a k |. We choose r n in such a way that lim n→∞ n(1 − r n ) = 1, e.g. r n = 1 − 1 n . Then lim n→∞ ε n+1 n(1−r n ) = 0 since lim n→∞ na n = 0. By Problem 18 (2), we have n k=0 |a k |(1 −r n )k = (1 −r n )n · n k=0 |a k |k n → 0, as n → ∞. Combined, we conclude lim n→∞ β n = 0. Remark 2. The result in this problem is a special case of the so-called Landau’s Theorem, see, for example, Boos [1], §5.1 Boundary behavior of power series. 2 Cauchy Integral Theorem and Cauchy Integral Formula 1. Proof. (i) The integral is equal to 2πi · sini = π(e −1 −e). (ii) The integral is equal to 2eπi. (iii) For sufficiently small ε > 0, the integral is equal to _ |z−π|=ε cos z (z +π)(z −π) dz + _ |z+π|=ε cos z (z +π)(z −π) dz = 2πi _ cos π 2π + cos(−π) −2π _ = 0. (iv) The integral is equal to 2πi _ 1 z−3 _ (n−1) ¸ ¸ ¸ ¸ z=1 = 2πi · −(n−1)! 2 n = − (n−1)! 2 n−1 πi. (v) For sufficiently small ε > 0he integral is equal to _ |z+i|=ε 1 (z −i)(z 2 + 4) · dz (z +i) + _ |z−i|=ε 1 (z +i)(z 2 + 4) · dz (z −i) = 2πi · _ 1 (−2i) · 3 + 1 2i · 3 _ = 0. (vi) By the Fundamental Theorem of Algebra, the equation z 5 −1 = 0 has five solutions z 1 , z 2 , z 3 , z 4 , z 5 . For any i ∈ {1, 2, 3, 4, 5}, we have (z i −z 1 ) · · · (z i −z i−1 )(z i −z i+1 ) · · · (z i −z 5 ) = lim z→z i z 5 −1 z −z i = lim z→z i z 5 −z 5 i z −z i = 5z 4 i = 5 z i . So for sufficiently small ε > 0, the integral is equal to 5 i=1 _ |z−z i |=ε dz (z −z 1 )(z −z 2 )(z −z 3 )(z −z 4 )(z −z 5 ) = 2πi 5 i=1 1 (z i −z 1 ) · · · (z i −z i−1 )(z i −z i+1 ) · · · (z i −z 5 ) = 2πi 6 i=1 z i 5 = 0, where the last equality comes from the fact that the expansion of (z−z 1 ) · · · (z−z 5 ) is z 5 −1 and −(z 1 +· · ·+z 5 ) is the coefficient of z 4 . (vii) If |a|, |b| > R, the integral is equal to 0. If |a| > R > |b|, the integral is equal to 2πi (b−a) n . If |b| > R > |a|, the integral is equal to 2πi _ 1 z−b _ (n−1) ¸ ¸ ¸ ¸ z=a = 2πi · (−1) n−1 (a − b) −n . If R > |a|, |b|, the integral is equal to 2πi _ 1 (b−a) n + (−1) n−1 1 (a−b) n _ = 0. 14 (viii) Denote by z 1 , z 2 , z 3 the three roots of the equation z 3 −1 = 0. Then the integral is equal to 2πi _ 1 (z 1 −z 2 )(z 1 −z 3 )(z 1 −2) 2 + 1 (z 2 −z 1 )(z 2 −z 3 )(z 2 −2) 2 + 1 (z 3 −z 1 )(z 3 −z 2 )(z 3 −2) 2 + _ 1 z 3 −1 _ (1) ¸ ¸ ¸ ¸ ¸ z=2 _ = 2πi _ z 1 3(z 1 −2) 2 + z 2 3(z 2 −2) 2 + z 3 3(z 3 −2) 2 − 12 49 _ . 2. Proof. 1 2πi _ C e zζ ζ n · dζ ζ = 1 2πi _ C ∞ k=0 z k k! 1 ζ n−k · dζ ζ = ∞ k=0 z k k! · 1 2πi _ C 1 ζ n−k · dζ ζ = z n n! , where the last equality is due to Cauchy’s integral formula. We need to justify the exchange of integration and infinite series summation. Indeed, for m > n, we have ¸ ¸ ¸ ¸ ¸ m k=0 z k k! · 1 2πi _ C 1 ζ n−k · dζ ζ − 1 2πi _ C ∞ k=0 z k k! 1 ζ n−k · dζ ζ ¸ ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ ¸ 1 2πi _ C ∞ k=m+1 z k k! 1 ζ n−k · dζ ζ ¸ ¸ ¸ ¸ ¸ ≤ 1 2π _ C ∞ k=m+1 |zζ| k k! |dζ| |ζ| n+1 ≤ 1 2π _ C ∞ k=m+1 M k k! |dζ| |ζ| n+1 , where M = |z| max ζ∈C |ζ|. Since ∞ k=m+1 M k k! is the remainder of e M , for m sufficiently large, ∞ k=m+1 M k k! can be smaller than any given positive number ε. So for m sufficiently large, ¸ ¸ ¸ ¸ ¸ m k=0 z k k! · 1 2πi _ C 1 ζ n−k · dζ ζ − 1 2πi _ C ∞ k=0 z k k! 1 ζ n−k · dζ ζ ¸ ¸ ¸ ¸ ¸ ≤ ε 1 2π _ C |dζ| |ζ| n+1 . This shows ∞ k=0 z k k! · 1 2πi _ C 1 ζ n−k · dζ ζ converges to 1 2πi _ C ∞ k=0 z k k! 1 ζ n−k · dζ ζ . 3. Proof. We note g(ζ) zζ−1 = g(ζ) ζ− 1 z . If |z| < 1, we apply Cauchy’s integral formula to 1 2πi _ |ζ|=1 f(ζ) ζ−z dζ and apply Cauchy’s integral theorem to 1 2πi _ |ζ|=1 g(ζ) ζ− 1 z dζ (regarding 1 z as a parameter and g(ζ) ζ− 1 z an analytic function of ζ). If |z| > 1, we apply Cauchy’s integral theorem to 1 2πi _ |ζ|=1 f(ζ) ζ−z dζ (regarding z as a parameter and f(ζ) ζ−z an analytic function of ζ) and apply Cauchy’s integral formula to 1 2πi _ |ζ|=1 g(ζ) ζ− 1 z dζ. 4. Proof. For z ∈ {z : |z| < 3}, by Cauchy’s integral formula, f(z) = 2πi(3z 3 + 7z + 1). So f (1 + i) = 2πi[6(1 +i) + 7] = −12π + 26πi. 5. 15 Proof. We note _ |z|=1 (z + 1 z ) 2n dz z = _ 2π 0 (e iθ +e −iθ ) 2n idθ = 2 2n i _ 2π 0 cos 2n θdθ. On the other hand, (z + 1 z ) 2n · 1 z = 2n k=0 C k 2n z k (z −1 ) 2n−k · z −1 = 2n k=0 C k 2n z −2n+2k−1 . Therefore _ 2π 0 cos 2n θdθ = 1 2 2n i _ |z|=1 (z + 1 z ) 2n dz z = 1 2 2n i _ |z|=1 C n 2n dz z = 2πC n 2n 2 2n = 2π · 1 · 3 · 5 · · · · · (2n −1) 2 · 4 · 6 · · · · · 2n . 6. Proof. This is straightforward from Cauchy’s integral formula. 7. Proof. For any R > 0, we have f (n) (0) = n! 2πi _ |z|=R f(ξ) ξ n+1 dξ. So ¸ ¸ ¸ ¸ f (n) (0) n! ¸ ¸ ¸ ¸ = 1 2π ¸ ¸ ¸ ¸ _ 2π 0 |f(Re iθ )| R n dθ ¸ ¸ ¸ ¸ ≤ 1 2π ¸ ¸ ¸ ¸ ¸ _ 2π 0 Me |Re iθ | R n dθ ¸ ¸ ¸ ¸ ¸ = M e R R n . Let R = n, we get the desired inequality. Remark 3. Note h(R) = ln e R R n = R − nlnR takes minimal value at n. This is the reason why we choose R = n, to get the best possible estimate. Another perspective is to assume f(z) = e z . Then ¸ ¸ ¸ f (n) (0) n! ¸ ¸ ¸ = 1 n! ∼ 1 √ 2πn( n e ) n = 1 √ 2πn _ e n _ n by Stirling’s formula. So the bound M e R R n is not “far” from the best one. 8. (1) Proof. For any z ∈ D 2 , we can find R > 0 sufficiently large, so that z is in the region enclosed by γ and {ζ : |ζ −z| = R}. By Cauchy’s integral formula f(z) = 1 2πi _ −γ f(ζ) ζ −z dζ + 1 2πi _ |ζ|=R f(ζ) ζ −z dζ. So 1 2πi _ γ f(ζ) ζ −z dζ = −f(z) + 1 2πi _ |ζ|=R f(ζ) ζ −z dζ = −f(z) + 1 2πi _ |ζ|=R f(ζ) −f(∞) ζ −z dζ + 1 2πi _ |ζ|=R f(∞) ζ −z dζ. We note ¸ ¸ ¸ ¸ ¸ 1 2πi _ |ζ|=R f(ζ) −f(∞) ζ −z dζ ¸ ¸ ¸ ¸ ¸ ≤ sup ζ∈D(z,R) |f(ζ) −A| → 0, as R → ∞, and Cauchy’s integral formula implies 1 2πi _ |ζ|=R f(∞) ζ −z dζ = f(∞) · 1 = A. 16 So by letting R → ∞ in the above equality, we have 1 2πi _ γ f(ζ) ζ −z dζ = −f(z) +A. If z ∈ D 1 , h(ζ) = f(ζ) ζ−z is a holomorphic function in D 2 , with z a parameter. So for R sufficiently large, Cauchy’s integral theorem implies 0 = 1 2πi _ −γ f(ζ) ζ −z dζ + 1 2πi _ |ζ|=R f(ζ) ζ −z dζ. An argument similar to (1) can show 1 2πi _ γ f(ζ) ζ−z dζ = A. (2) Proof. We note z zζ−ζ 2 = 1 ζ − 1 ζ−z . For R sufficiently large, Cauchy’s integral theorem implies 0 = 1 2πi _ −γ f(ζ) ζ dζ + 1 2πi _ |ζ|=R f(ζ) ζ dζ. So 1 2πi _ γ f(ζ) ζ dζ = 1 2πi _ |ζ|=R f(ζ) ζ dζ. Similarly, 1 2πi _ γ f(ζ) ζ −z dζ = _ 1 2πi _ |ζ|=R f(ζ) ζ−z dζ −f(z) z ∈ D 2 1 2πi _ |ζ|=R f(ζ) ζ−z dζ z ∈ D 1 . Therefore 1 2πi _ γ zf(ζ) zζ −ζ 2 dζ = 1 2πi _ γ _ f(ζ) ζ − f(ζ) ζ −z _ dζ = _ f(z) z ∈ D 2 0 z ∈ D 1 . 9. Proof. Since f(z) = 0 in D, 1/f(z) is holomorphic in D. Applying Maximum Modulus Principle to 1/f(z), we conclude |f(z)| achieves its minimum on ∂D. Applying Maximum Modulus Principle to f(z), we conclude |f(z)| achieves its maximum on ∂D. Since |f(z)| ≡ M on ∂D, |f(z)| must be a constant on ¯ D. By Chapter 1, exercise problem 21 (iii), we conclude f(z) is a constant function. 10. Proof. For R large enough, |a|, |b| < R. So we can find ε > 0 such that (assume a = b) _ |z|=R f(z) (z −a)(z −b) dz = _ |z−a|=ε f(z) (z −a)(z −b) dz + _ |z−b|=ε f(z) (z −a)(z −b) dz = 2πi a −b [f(a) −f(b)]. Meanwhile, denote by M a bound of f, we have ¸ ¸ ¸ ¸ ¸ _ |z|=R f(z) (z −a)(z −b) dz ¸ ¸ ¸ ¸ ¸ ≤ M _ |z|=R |dz| |(z −a)(z −b)| ≤ M _ |z|=R |dz| (R −|a|)(R −|b|) = 2πMR (R −|a|)(R −|b|) → 0, as R → ∞. Combined, we conclude 0 = 2πi a−b [f(a) −f(b)], ∀a, b ∈ C. This shows f is a constant function. 17 11. Proof. By Cauchy’s integral formula, |f (n) (0)| ≤ n! 2π ¸ ¸ ¸ ¸ ¸ _ |z|=r f(ξ) ξ n+1 dξ ¸ ¸ ¸ ¸ ¸ ≤ n! 2π _ 2π 0 1 1−r r n dθ = n! r n (1 −r) . 12. Proof. We note lim z→z 0 Φ(z) −Φ(z 0 ) z −z 0 = 1 2πi lim z→z 0 _ γ ϕ(ζ) (ζ −z)(ζ −z 0 ) dζ = 1 2πi _ γ ϕ(ζ) (ζ −z 0 ) 2 dζ, where the exchange of limit and integration can be seen as an application of Lebesgue’s Dominated Conver- gence Theorem (We can find ρ > 0 so that ∀z in a neighborhood of z 0 , dist(γ, z) ≥ ρ. Then 1 (ζ−z)(ζ−z 0 ) ) ≤ 1 ρ 2 ). This shows Φ is holomorphic in any region D which does not contain any point of γ. The formula for n-th derivative of Φ can be proven similarly and by induction. For details, we refer the reader to Fang [3], Chapter 4, §3, Lemma 2. Remark 4. The result still holds if ϕ is piecewise continuous on γ. See, for example, Fang [3], Chapter 7, proof of Theorem 7. 13. Proof. Since f(z) = 0 in D, 1/f(z) is holomorphic in D. Applying Maximum Modulus Principle to 1/f(z), we conclude |f(z)| does not achieve its minimum in D unless it’s a constant function. Applying Maximum Modulus Principle to f(z), we conclude f(z) does not achieve its maximum in D unless it’s a constant function. Combined, we have m < |f(z)| < M for any point z ∈ D. 14. Proof. Define f(z) = p n (z) z n and g(z) = f( 1 z ). Then g(z) is holomorphic and bounded on D(0, 1) \ {0}, since lim z→0 g(z) = lim z→∞ p n (z) z n is a finite number and g(z) is clearly continuous in a neighborhood of ∂D(0, 1). By Theorem 2.11, Riemann’s Theorem, g(z) can be analytically continued to D(0, 1). By a corollary of Theorem 2.18, Maximum Modulus Principle, the maximum value of |g(z)| can only be reached on ∂D(0, 1). So |g(z)| ≤ max |z|=1 |g(z)| = max |z|=1 |f(z)| = max |z|=1 |p n (z)| ≤ M, ∀z ∈ D(0, 1), i.e. |f(z)| = |p n (z)| |z n | ≤ M for 1 ≤ |z| < ∞. 15. Proof. Define g(z) = f(Rz) M (z ∈ D(0, 1)) and apply Theorem 2.19, Schwarz Lemma, to g(z), we have |g(z)| ≤ |z| (∀z ∈ D(0, 1)) and |g (0)| ≤ 1. So for any z ∈ D(0, R), f(z) = f(R · z R ) = Mg( z R ) ≤ M |z| R and |f (0)| = |g (0)| · M R ≤ M R . 16. Proof. Define ρ = dist(C \ V, K). Then ρ ∈ (0, ∞). By Theorem 2.7, for any z ∈ K, we have |f (n) (z)| ≤ n! 2π _ ∂V |f(ξ)| |ξ −z| n+1 |dξ| ≤ n! 2π _ ∂V |dξ| ρ n+1 · sup z∈V |f(z)|. Define c n = n! 2π _ ∂V |dξ| ρ n+1 and take supremum on the left side of the above inequality, we have sup z∈K |f (n) (z)| ≤ c n sup z∈V |f(z)|. 18 17. Proof. We have to assume γ has finite length, that is, L(γ) := _ γ |dz| < ∞. Then by the definition of uniform convergence, for any ε > 0, there exists n 0 such that ∀n ≥ n 0 , sup z∈D |f(z) − f n (z)| ≤ ε L(γ) . So for any n ≥ n 0 , ¸ ¸ ¸ ¸ _ γ f(z)dz − _ γ f n (z)dz ¸ ¸ ¸ ¸ ≤ _ γ |f(z) −f n (z)||dz| ≤ _ γ ε L(γ) |dz| = ε. Therefore lim n→∞ _ γ f n (z)dz = _ γ f(z)dz. 18. Proof. We note the linear fractional transformation ω(z) = z−α z+α maps {z : Rez > 0} to D(0, 1). So h(z) = ω(f(z)) maps D(0, 1) to itself (see, for example, Fang [3], Chapter 3, §7) and h(0) = 0. By Schwarz Lemma, h(z)| = |z| and |h (0)| ≤ 1. This is equivalent to ¸ ¸ ¸ ¸ f(z) −α f(z) +α ¸ ¸ ¸ ¸ ≤ |z|, and |f (0)| ≤ 2α. 19. Proof. The linear fractional transformation ω(z) = z z−2A maps {z : Rez < A} to D(0, 1). So h(z) = ω(f(z)) = f(z) f(z)−2A maps D(0, 1) to itself and h(0) = 0. By Schwarz Lemma, |h(z)| ≤ |z|. So ∀z ∈ D(0, 1) |f(z)| ≤ |z| · |f(z) −2A| ≤ |z||f(z)| + 2A|z|, i.e. |f(z)| ≤ 2A|f(z)| 1 −|z| . 20. (1) Proof. We note e z +e −z + 2 cos z 4 = 1 4 [e z +e −z +e iz +e −iz ] = 1 4 ∞ k=0 1 k! [z k + (−z) k + (iz) k + (−iz) k ] = ∞ m=0 z 4m (4m)! . The radius of convergence is ∞. (2) Proof. By induction, it is easy to show _ 1 (1−ζ) 3 _ (n) = (n+2)! 2! (1 −ζ) −(n+3) . So the Taylor expansion of 1 (1−ζ) 3 is 1 (1−ζ) 3 = ∞ n=0 (n+1)(n+2) 2 ζ n , with radius of convergence equal to 1. This implies ζ + 4ζ 2 +ζ 3 (1 −ζ) 3 = ζ + 7ζ 4 + ∞ n=3 (3n 2 −3n + 1)ζ n . Replace ζ with z 2 , we can get the Taylor expansion at z = 0: z 2 + 7z 8 + ∞ n=3 (3n 2 − 3n + 1)z 2n , and the radius of convergence is equal to 1 (this can be verified by calculating lim n→∞ (3n 2 −3n + 1) 1 n = 1). (3) 19 Proof. By induction, it is easy to show _ (1 −ζ) −4 ¸ (n) = (n+3)! 3! (1 −ζ) −(n+4) . So for ζ ∈ D(0, 1), (1 −ζ) −4 = ∞ n=0 (n+1)(n+2)(n+3) 6 ζ n . Replace ζ with z 5 , we get for z ∈ D(0, 1) 1 (1 −z 5 ) 4 = ∞ n=0 (n + 1)(n + 2)(n + 3) 6 z 5n . Therefore (1 −z −5 ) −4 = z 20 (1 −z 5 ) 4 = ∞ n=0 (n + 1)(n + 2)(n + 3) 6 z 5n+20 . By Hadamard’s formula, the radius of convergence is 1. (4) Proof. By induction, we can show _ (1 +z) −2 ¸ (n) = (−1) n (n + 1)!(1 +z) −(n+2) . So ∀z ∈ D(0, 1), 1 (1 +z) 2 = ∞ n=0 (−1) n (n + 1)z n . It is also easy to verify that 1 (z + 1) 2 (z −1) = − 1 4(z + 1) + 1 4(z −1) − 1 2(z + 1) 2 = − 1 4 ∞ n=0 (−z) n − 1 4 ∞ n=0 z n − 1 2 ∞ n=0 (−1) n (n + 1)z n . Therefore z 6 (z + 1)(z 2 −1) = − z 6 4 ∞ n=0 [(−1) n + 1 + 2 · (−1) n (n + 1)] z n = − 1 4 ∞ n=0 [(−1) n (2n + 3) + 1] z n+6 . By Hadamard’s formula, the radius of convergence is 1. 21. (1) Proof. We note 1 2π _ π −π |f(re iθ )| 2 dθ = 1 2π _ π −π _ ∞ n=0 a n r n e inθ _ · _ ∞ m=0 ¯ a m r m e −imθ _ dθ = 1 2π ∞ m,n=0 _ π −π a n ¯ a m r n+m e i(n−m)θ dθ = ∞ m,n=0 a n ¯ a m r n+m δ nm = ∞ n=0 |a n | 2 r 2n , where δ mn = _ 1 if m = n 0 if m = n and the exchange of integration and summation is justified by Theorem 1.14 (Abel Theorem) and Problem 17. (2) 20 Proof. In the result of part (1), multiply both sides by r and integrate with respect to r from 0 to 1, we have ∞ n=0 |a n | 2 _ 1 0 r 2n+1 dr = ∞ n=0 |a n | 2 2n + 2 = 1 2π _ 1 0 _ π −π |f(re iθ )| 2 rdrdθ = 1 2π _ _ D |f(z)| 2 dA. So ∞ n=0 |a n | 2 n+1 = 1 π _ _ D |f(z)| 2 dA. 22. Proof. Suppose u is harmonic on D(0, R) and continuous ¯ D(0, R), then ∀z ∈ ¯ D(0, 1), v(z) := u(zR) is harmonic on D(0, 1) and continuous on ¯ D(0, 1). By formula (2.31), for any z ∈ D(0, R), u(z) = v( z R ) = 1 2π _ 2π 0 v(e iτ ) 1 − ¸ ¸ z R ¸ ¸ 2 ¸ ¸ 1 − ¯ z R e iτ ¸ ¸ 2 dτ = 1 2π _ 2π 0 u(Re iτ ) R 2 −|z| 2 |R − ¯ ze iτ | 2 dτ = 1 2π _ 2π 0 u(Re iτ ) R 2 −|z| 2 |R −ze −iτ | 2 dτ = 1 2π _ 2π 0 u(ζ) R 2 −|z| 2 |ζ −z| 2 dτ. This is formula (2.32). To verify formula (2.33), note Re Re iτ +z Re iτ −z = Re (Re iτ +z)(Re −iτ − ¯ z) |Re iτ −z| 2 = Re R 2 −|z| 2 +R(ze −iτ − ¯ ze iτ ) |Re iτ −z| 2 = R 2 −|z| 2 |Re iτ −z| 2 . Therefore, u(z) = 1 2π _ 2π 0 u(Re iτ ) R 2 −|z| 2 |Re iτ −z| 2 dτ = 1 2π _ 2π 0 u(Re iτ )Re Re iτ +z Re iτ −z dτ = Re _ 1 2π _ 2π 0 u(Re iτ ) Re iτ +z Re iτ −z dτ _ = Re _ 1 2πi _ |ζ|=R u(ζ) ζ +z ζ −z dζ ζ _ . This is formula (2.33). 23. Proof. On ∂D(0, 1), |(−6z)−(z 4 −6z +3)| = |z 4 +3| ≤ 4 < | −6z|. By Rouch´e Theorem, z 4 −6z +3 and −6z have the same number of zeros in D(0, 1), which is one. On ∂D(0, 2), |z 4 −(z 4 −6z+3)| = |6z−3| ≤ 15 < |z 4 |. So z 4 − 6z + 3 and z 4 have the same number of zeros in D(0, 2), which is four. Combined, we conclude z 4 −6z + 3 = 0 has one root in D(0, 1) and three roots in the annulus {z : 1 < |z| < 2}. 24. Proof. On ∂D(0, 1), |(−5z 4 ) −(z 7 −5z 4 −z +2)| = |z 7 −z +2| ≤ 4 < | −5z 4 |. So z 7 −5z 4 −z +2 and −5z 4 have the same number of zeros in D(0, 1), which is four (counting multiplicity). 25. Proof. Let P(z) = z 4 + 2z 3 − 2z + 10. We note P(z) = (z 2 − 1)(z + 1) 2 + 11. If z ∈ R and |z| ≥ 1, P(z) ≥ 11; if z ∈ R and |z| ≤ 1, P(z) ≥ −1 · (1 + 1) 2 + 11 = 7. So P(z) = 0 has no root on the real axis. If z = iy with y ∈ R, we have P(iy) = y 4 + 10 − 2iy(y 2 + 1) = 0. So P(z) = 0 has no root on the imaginary axis. Consider the region D enclosed by the curve γ = γ 1 ∪γ 2 ∪γ 3 , where (R is a positive number) γ 1 = {z : 0 ≤ Rez ≤ R, Imz = 0}, γ 2 = {z : |z| = R, arg z ∈ [0, π 2 ]}, and γ 3 = {z : 0 ≤ Imz ≤ R, Rez = 0}. On γ 1 , ∆ γ 1 arg P(z) = 0. On γ 2 , P(z) = z 4 _ 1 + 2z 3 −2z+10 z 4 _ . So ∆ γ 2 arg P(z) = 4 · π 2 +o(1) = 2π +o(1) (R → ∞). On γ 3 , ∆ γ 3 arg P(z) = arg P(0) − arg P(iR) = arg 10 − arg(R 4 + 10 − 2iR(R 2 + 1)) = 0 − arg _ 1 −2i R(R 2 +1) R 4 +10 _ = o(1) (R → ∞). Combined, we have ∆ γ arg P(z) = 3 k=1 ∆ γ k arg P(z) = 2π + o(1) (R → ∞). 21 By The Argument Principle, P(z) = 0 has only one root in the first quadrant. The root conjugate to this root must lie in the fourth quadrant. The other two roots are conjugate to each other and are located in the second and third quadrant. So one must lie in the second quadrant and the other must lie in the third quadrant. Remark 5. The above solution can be found in Fang [3], Chapter 6, §3 Example 5. 26. Proof. On ∂D(0, 1), |(−8z +10) −(z 4 −8z +10)| = |z 4 | = 1 < |10 −8|z|| ≤ | −8z +10|. So z 4 −8z +10 and −8z + 10 have the same number of zeros in D(0, 1), which is zero. On ∂D(0, 3), |z 4 −(z 4 −8z +10)| = |8z −10| < 34 < |z 4 |. So z 4 −8z +10 and z 4 have the same number of zeros in D(0, 3), which is four (counting multiplicity). So z 4 −8z +10 = 0 has no root in D(0, 1) and four roots in the annulus {z : 1 < |z| < 3}. 27. Proof. If a > e, on ∂D(0, 1), |az n −(az n −e z )| = |e z | ≤ e |z| = e < |az n |. By Rouch´e Theorem, e z −az n and az n have the same number of zeros in D(0, 1), which is n. 28. Proof. On ∂D(0, 1), |z −(z −f(z))| = |f(z)| < 1 = |z|. So z −f(z) and z have the same number of zeros in D(0, 1), which is one. So there is a unique fixed point of f in D(0, 1). Remark 6. This is a special case of Brower’s Fixed Point Theorem. 29. (1) Proof. Re(ze z ) = e x (xcos y −y siny). (2) Proof. Let u(x, y) = ax 3 +bx 2 y +cxy 2 +dy 3 . Then ∆u = (6a + 2c)x + (2b + 6d)y. So u is harmonic if and only if 3a+c = 0 and b +3d = 0. So the most general harmonic function of the form ax 3 +bx 2 y +cxy 2 +dy 3 is ax 3 −3dx 2 y −3axy 2 +dy 3 . 30. Proof. Let u(r, θ) = ln(1−2r cos θ +r 2 ). Then using the fact ∆ = 1 r ∂ ∂r _ r ∂ ∂r _ + 1 r 2 ∂ 2 ∂θ 2 , we can verify ∆u = 0. So by the mean-value property of harmonic functions, we have 1 2π _ 2π 0 u(r, θ)dθ = u(0, 0) = 0. Since cos θ is an even function of θ, it’s not hard to show _ 2π 0 ln(1 − 2r cos θ + r 2 )dθ = 2 _ π 0 ln(1 − 2r cos θ + r 2 )dθ. So _ π 0 ln(1 −2r cos θ +r 2 )dθ = 0. 31. Proof. Let M = sup z∈C |u(z)|. Then ∀z ∈ C and R > |z|, the holomorphic function defined in formula (2.34) satisfies |f(z)| ≤ 1 2π _ 2π 0 |u(Re iθ )| ¸ ¸ ¸ ¸ Re iθ +z Re iθ −z ¸ ¸ ¸ ¸ dθ ≤ M 2π _ 2π 0 R +|z| R −|z| dθ → M, as R → ∞. So f(z) is bounded on the entire complex plane, and by Liouville’s Theorem, f is a constant function. Therefore u = Re(f) is a constant function. 32. 22 Proof. If the arc is ∂D(0, 1), the desired harmonic function u ≡ 1. So without loss of generality, we assume the arc γ = {z ∈ ∂D(0, 1) : θ 1 ≤ arg z ≤ θ 2 } with 0 < θ 2 −θ 1 < 2π. For n sufficiently large, we consider γ n = {z ∈ ∂D(0, 1) : θ 1 − 1 n ≤ arg z ≤ θ 2 + 1 n }. By the Partition of Unity Theorem for R 1 , we can find ϕ n ∈ C(∂D(0, 1)) such that 0 ≤ ϕ n ≤ 1, ϕ n ≡ 1 on γ, and ϕ n ≡ 0 on ∂D(0, 1) \ γ n . Define u n (z) = _ 2π 0 P(ζ, z)ϕ n (ζ)dθ where P(·, ·) is the Poisson kernel and ζ = Re iθ . Then u n is the unique solution of the Dirichlet problem with boundary value ϕ n . So ϕ n is harmonic in D(0, 1). It is easy to see that if (ϕ n ) n uniformly converges to the indicator function 1 γ (z), (u n ) n uniformly converges to u(z) := _ 2π 0 P(ζ, z)1 γ (ζ)dθ on D(0, ρ) (ρ ∈ (0, 1)). Since each u n satisfies the local mean value property, u must also satisfies the local mean value property in D(0, 1) and is continuous. By Theorem 2.22, u is harmonic on D(0, 1). This suggests the harmonic function we are looking for is exactly u. What is left to prove is that for any z 0 ∈ ∂D(0, 1), lim |z|=1,z→z 0 u(z) = 1 γ (z 0 ). Indeed, in general, we have the following classical result: Suppose function ϕ is piecewise continuous on ∂D(0, 1), then the function u(z) := _ 2π 0 P(ζ, z)ϕ(ζ)dθ is harmonic in D(0, 1) and for any continuity point z 0 of ϕ, we have lim z→z 0 ,|z|<1 u(z) = ϕ(z 0 ). For the proof of continuity on the boundary, we refer to Fang [3] Chapter 7, Theorem 7. 33. Proof. ∀ε > 0, there exists N, such that for any n ≥ N, p ∈ N, we have sup ζ∈∂U ¸ ¸ ¸ ¸ ¸ n+p k=n+1 f k (ζ) ¸ ¸ ¸ ¸ ¸ < ε. By Maximum Modulus Principle, sup z∈ ¯ U ¸ ¸ ¸ ¸ ¸ n+p k=n+1 f k (z) ¸ ¸ ¸ ¸ ¸ ≤ sup ζ∈∂U ¸ ¸ ¸ ¸ ¸ n+p k=n+1 f k (ζ) ¸ ¸ ¸ ¸ ¸ < ε. So Cauchy criterion implies ∞ n=1 f n (z) converges uniformly on ¯ U. Remark 7. This result is the so-called Weierstrass’ Second Theorem. 34. Proof. Define g(z) = f(z)−f(z 0 ) z−z 0 = f(z) z−z 0 , ∀z ∈ ¯ D(0, R) \ {z 0 }. Then g(z) can be analytically continued to D(0, R) (Theorem 2.11, Riemann Theorem) and is therefore continuous on ¯ D(0, R). By Cauchy’s integral formula, we have f(0) −z 0 = g(0) = 1 2πi _ |ζ|=R g(ζ) ζ dζ = 1 2πi _ |ζ|=R f(ζ) ζ(ζ −z 0 ) dζ. Therefore ¸ ¸ ¸ ¸ f(0) z 0 ¸ ¸ ¸ ¸ ≤ 1 2π _ 2π 0 |f(Re iθ ) |Re iθ −z 0 | dθ ≤ M R −|z 0 | , which is equivalent to R|f(0)| ≤ (M +|f(0)|)|z 0 |. Remark 8. The point of this problem is to give an estimate of a holomorphic function’s modulus at z = 0 in terms of its zeros. 35. Proof. This problem seems suspicious. For example, when z 0 = 0, of course n k=1 |z 0 −z k | = n k=1 |z k | > 1. When z 0 is sufficiently large, it is also clear n k=1 |z 0 − z k | > 1. So the point z 0 can be both inside and outside D(0, 1). I’m not sure what proof is needed for such a trivial claim. 23 36. Proof. By Maximum Modulus Principle, M(r) = max |z|≤r |f(z)|. So M(r) is an increasing function on [0, R). 37. Proof. Suppose there is an n-th degree polynomial p(z) = a 0 +a 1 z +· · · +a n z n such that n ≥ 1, a n = 0, and p(z) = 0 for any z ∈ C. Then q(z) = 1/p(z) is holomorphic over the whole complex plane. For any R > 0, we have by Maximum Modulus Principle max |z|≤R |q(z)| ≤ max |z|=R |q(z)|. Since on {z : |z| = R}, |p(z)| ≥ |a n |R n −|a n−1 |R n−1 −· · ·−|a 1 |R−|a 0 | = R n _ |a n | − |a n−1 | R −· · · − |a 1 | R n−1 − |a 0 | R n _ → ∞ as R → ∞, we must have lim R→∞ max |z|=R |q(z)| = 0. This implies sup z∈C |q(z)| = 0, i.e. p(z) ≡ ∞. This is a contradiction and our assumption must be incorrect. 38. Proof. Apply the Maximum Modulus Principle to the function g(z) = 1/f(z), which is holomorphic on U. 39. Proof. Note the Laplace operator ∆ = 4 ∂ ¯ ∂ ∂ ∂ , so log |z| is a harmonic function on C \ {0} and log |f(z)| is a harmonic function outside the zeros of f(z). Define K = {z ∈ U : f(z) = 0} and V ε = ∪ z∈K D(z, ε). Then for any α ∈ R, αlog |z| + log |f(z)| is harmonic in U \ ¯ V ε and continuous on ¯ U \ V ε . By Maximum Modulus Principle for harmonic functions, max z∈ ¯ U\V ε (αlog |z| + log |f(z)|) = max z∈∂( ¯ U\V ε ) (αlog |z| + log |f(z)|). When z approaches to zeros of f(z), log |f(z)| → −∞, so by letting ε → 0, we can further deduce that max z∈U (αlog |z| + log |f(z)|) = max z∈∂U (αlog |z| + log |f(z)|). Therefore αlog |z| + log |f(z)| ≤ max{αlog r 1 + log M(r 1 ), αlog r 2 + log M(r 2 )}, ∀z ∈ U, which is the same as αlog r + log M(r) ≤ max{αlog r 1 + log M(r 1 ), αlog r 2 + log M(r 2 )}, r ∈ [r 1 , r 2 ]. Now let α be such that the two values inside the parentheses on the right are equal, that is α = log M(r 2 ) −log M(r 1 ) log r 1 −log r 2 . Then from the previous inequality, we get log M(r) ≤ αlog r 1 + log M(r 1 ) −αlog r, which upon substituting value for α gives log M(r) ≤ (1 −s) log M(r 1 ) +s log M(r 2 ), where s = log r 1 −log r log r 2 −log r 1 . 40. 24 Proof. Fix r ∈ (0, 1). For any ρ ∈ (0, r) and any z ∈ D(0, ρ), we have z n ∈ D(0, ρ) (∀n ∈ N). By Cachy’s integral formula, ∀n, p ∈ N, we have n+p k=n+1 |f(z k )| = n+p k=n+1 |f(z k ) −f(0)| = n+p k=n+1 1 2π ¸ ¸ ¸ ¸ ¸ _ ∂D(0,r) _ f(ζ) ζ −z k − f(ζ) ζ _ dζ ¸ ¸ ¸ ¸ ¸ ≤ n+p k=n+1 1 2π ¸ ¸ ¸ ¸ ¸ _ ∂D(0,r) f(ζ) ζ(ζ −z k ) dζ · z k ¸ ¸ ¸ ¸ ¸ ≤ n+p k=n+1 1 2π _ 2π 0 |f(re iθ )|dθ r −ρ k · ρ k ≤ 1 2π _ 2π 0 |f(re iθ )|dθ r −ρ n+p k=n+1 ρ k → 0 as n → ∞. So ∞ n=1 f(z n ) converges absolutely and uniformly on ¯ D(0, ρ). Remark 9. In general, there is no Mean Value Theorem for holomorphic functions. For example, f(z) = exp _ i 2z−(z 1 +z 2 ) z 2 −z 1 π _ is analytic but f(z 2 ) − f(z 1 ) = f (z)(z 2 − z 1 ), ∀z ∈ C (for more details, see Qazi [5]). But as the proof of Theorem 2.7 shows, the formula f(z) −f(z 0 ) = z −z 0 2πi _ ∂U f(ζ)dζ (ζ −z)(ζ −z 0 ) more or less fills the void. In particular, it shows holomorphic functions are locally Lipschitz. 41. Proof. This problem is the same as Problem 15. 42. Proof. This problem is the same as Problem 19. 43. (1) Proof. By Problem 18, ¸ ¸ ¸ f(z)−1 f(z)+1 ¸ ¸ ¸ ≤ |z|. So |f(z)| −1 ≤ |f(z)−1| ≤ |z||f(z)+1| ≤ |z||f(z)| +|z|. Since |z| < 1, the above inequality implies |f(z)| ≤ 1+|z| 1−|z| . From the same inequality ¸ ¸ ¸ f(z)−1 f(z)+1 ¸ ¸ ¸ ≤ |z|, we have (Ref(z) −1) 2 + (Imf(z)) 2 ≤ |z| 2 [(Ref(z) + 1) 2 + (Imf(z)) 2 ] ≤ |z| 2 (Ref(z) + 1) 2 + (Imf(z)) 2 . So |Ref(z) −1| ≤ |z|(Ref(z) +1). If Ref(z) > 1, we can deduce from this inequality 1−|z| 1+|z| ≤ 1 < Ref(z). If Ref(z) ≤ 1, we can deduce from this inequality 1 −Ref(z) ≤ |z|Ref(z) +|z|, i.e. 1−|z| 1+|z| ≤ Ref(z). (2) Proof. We note f(z) = 1+e iθ z 1−e iθ z if and only if f(z)−1 f(z)+1 = e iθ . So the claim is straightforward from Schwarz Lemma. 44. 25 Proof. Because D(0, 1) is pre-compact, by Theorem 1.11 (Bolzano-Weierstrass Theorem), it suffices to find a sequence (z n ) ∞ n=1 ⊂ D(0, 1), such that lim n→∞ z n = z 0 ∈ ∂D(0, 1) and (f(z n )) ∞ n=1 is bounded. Assume this does not hold. Then ∀z 0 ∈ ∂D(0, 1) and ∀n ∈ N, there exists δ(z 0 ) > 0 such that ∀z ∈ D(z 0 , δ(z 0 )) ∩D(0, 1), |f(z)| ≥ n. The family of these open sets (D(z, δ(z)) z∈∂D(0,1) is an open covering of the compact set ∂D(0, 1). By Theorem 1.10 (Heine-Borel Theorem), there is a finite sub-covering (D(z k , δ(z k )) p k=1 of ∂D(0, 1). Consequently, for each n ∈ N, we can find ε n > 0 such that ∀z ∈ {z : 1 −ε n ≤ |z| < 1}, |f(z)| ≥ n. Without loss of generality, we can assume (ε n ) ∞ n=1 monotonically decreases to 0. Since f(z) uniformly approaches to ∞ as z → ∂D(0, 1), by Theorem 2.13, f has finitely many zeros in D(0, 1). So f can be written as f(z) = m i=1 (z −z i ) k i · h(z), where h is a holomorphic function on ∂D(0, 1) with no zeros in D(0, 1). So h satisfies the Minimum Modulus Principle: ∀ρ > 0, |h(z)| ≥ min |ζ|=ρ |h(ζ)| for any z ∈ D(0, ρ). Therefore, for any z ∈ D(0, 1 −ε n ), |h(z)| ≥ min |ζ|=1−ε n |h(ζ)| = min |ζ|=1−ε n |f(ζ)| | m i=1 (ζ −z i ) k i | ≥ n max |ζ|=1−ε n m i=1 |ζ −z i | k i ≥ n 2 m i=1 k i . Fix z ∈ D(0, 1 −ε n ) and let n → ∞, we get h(z) = ∞. Contradiction. 45. Proof. By assumption, we can write f(z) as f(z) = n j=1 (z − z j ) k j · h(z), where h(z) is a holomorphic function in D(0, 1) with no zeros in D(0, 1). Define G(z) = h(z) · n j=1 (1 − ¯ z j z) k j . Then G is holomorphic on D(0, 1). Apply Maximum Modulus Principle to G(z) on {z : |z| ≤ 1 −ε} (ε > 0), we get |G(z)| ≤ max |z|=1−ε |h(z)| n j=1 |1 − ¯ z j z| k j = max |z|=1−ε |f(z)| n j=1 ¸ ¸ ¸ z−z j 1−¯ z j z ¸ ¸ ¸ k j ≤ max |z|=1−ε 1 n j=1 ¸ ¸ ¸ z−z j 1−¯ z j z ¸ ¸ ¸ k j . Since each z−z j 1−¯ z j z maps ∂D(0, 1) to ∂D(0, 1), by letting ε → 0, we get G(z) ≤ 1, i.e. |f(z)| ≤ n j=1 ¸ ¸ ¸ z−z j 1−¯ z j z ¸ ¸ ¸ k j . 46. Proof. Let a = f(0) and define ϕ a (ζ) = −ζ+a 1−¯ aζ . Then h(z) = ϕ a (f(z)) maps D(0, 1) to D(0, 1) with h(0) = ϕ a (a) = 0. So Schwarz Lemma implies |h (0)| ≤ 1. We note h (z) = ϕ a (f(z)) · f (z). Since ϕ a (ζ) = −(1 − ¯ aζ) −(−ζ +a)(−¯ a) (1 − ¯ aζ) 2 = −1 +|a| 2 (1 − ¯ aζ) 2 , we have h (0) = ϕ a (a)f (0) = f (0) |a| 2 −1 . So we have |f (0)| 1 −|a| 2 ≤ 1, i.e. |f (0)| ≤ 1 −|a| 2 . 48. Proof. The M¨obius transformation w(z) = z−i z+i maps C + to D = D(0, 1). So f ∈ Aut(C + ) if and only if w ◦ f ◦ w −1 ∈ Aut(D). By Theorem 2.20, ∃a ∈ D and τ ∈ R, such that w ◦ f ◦ w −1 (ζ) = ϕ a ◦ ρ τ (ζ). Plain calculation shows f(z) = w −1 ◦ ϕ a ◦ ρ τ ◦ w(z) = (1 +a)(z +i) −(1 + ¯ a)e iτ (z −i) (1 −a)(z +i) −(¯ a −1)e iτ (z −i) . 50. Proof. This problem is the same as Problem 10. 26 3 Theory of Series of Weierstrass Throughout this chapter, all the integration paths will take the following convention on orientation: all the arcs take counterclockwise as their orientation and all the segments take the natural orientation of R 1 as their orientation. 1. Proof. This is just Mittag-Leffler Theorem (Theorem 3.8). 2. Proof. To prove Theorem 3.1, choose varepsilon > 0 sufficiently small so that the circles {z : |z −z k | = ε} (1 ≤ k ≤ n) don’t intersect with each other or with ∂U. Then by Cauchy’s integration theorem, _ ∂U f(z)dz − n k=1 _ |z−z k |=ε f(z)dz = 0, which is _ ∂U f(z)dz = 2πi n k=1 Res(f, z k ). To prove Theorem 3.12, we choose R > 0 sufficiently large so that ¯ U ⊂ D(0, R). Then _ ∂U f(z)dz = 2πi n k=1 Res(f, z k ) and _ ∂D(0,R) f(z)dz − _ ∂U f(z)dz = 0. Since _ ∂D(0,R) f(z)dz = −2πiRes(f, ∞), we conclude n k=1 Res(f, z k ) + Res(f, ∞) = 0. 3. (i) Proof. Suppose 1 z 3 (z+i) has Laurent series ∞ n=−∞ c n (z + i) n . Then by Theorem 3.2 and Cauchy integral theorem, for ε ∈ (0, 1), c n = 1 2πi _ |ζ+i|=ε 1 (ζ +i) n+1 · 1 ζ 3 (ζ +i) dζ = 1 (n + 1)! d n+1 dζ n+1 (ζ −3 ) ¸ ¸ ¸ ¸ ζ=−i = 1 (n + 1)! (−1) n+1 (n + 3)! 2! (−i) −(n+4) = (−1) n+1 (n + 2)(n + 3) 2 i n . (ii) Proof. We note z 2 (z + 1)(z + 2) = (z + 1)(z + 2) −3z −2 (z + 1)(z + 2) = 1 − 3(z + 1) −1 (z + 1)(z + 2) = 1 + 1 z + 1 − 4 z + 2 = 1 + 1 z · 1 1 + 1 z −2 · 1 1 + z 2 = ∞ k=0 (−1) k z k+1 −1 − ∞ k=1 (−1) k 2 k−1 z k . (iii) Proof. We note log _ z −a z −b _ = log _ 1 − a z 1 − b z _ = log _ 1 − a z _ −log _ 1 − b z _ = − ∞ n=1 1 n _ a z _ n + ∞ n=1 1 n _ b z _ n = ∞ n=1 1 n (b n −a n )z −n . 27 (iv) Proof. We note z 2 e 1 z = z 2 ∞ k=0 z −k k! = z 2 +z + 1 2 + ∞ k=1 z −k (k + 2)! . (v) Proof. We note sinz = ∞ k=0 (−1) k z 2k+1 (2k+1)! . So sin z z+1 = ∞ k=0 (−1) k (2k+1)! z 2k+1 (z+1) 2k+1 . Suppose sin z z+1 has Laurent series ∞ n=−∞ c n (z + 1) n . Then by Theorem 3.2, for ε > 0, c n = 1 2πi _ |ζ+1|=ε 1 (ζ + 1) n+1 ∞ k=0 (−1) k (2k + 1)! ζ 2k+1 (ζ + 1) 2k+1 dζ = ∞ k=0 (−1) k (2k + 1)! · 1 2πi _ |ζ+1|=ε ζ 2k+1 (ζ + 1) 2k+n+2 dζ = k≥ −n−1 2 (−1) k (2k + 1)! 1 (2k +n + 1)! d 2k+n+1 dζ 2k+n+1 _ ζ 2k+1 _¸ ¸ ζ=−1 . Since d 2k+n+1 dζ 2k+n+1 _ ζ 2k+1 _¸ ¸ ζ=−1 = _ ¸ _ ¸ _ 0 n ≥ 1, (2k + 1)! n = 0, (2k+1)! (−n)! (−1) −n n ≤ −1, we have c n = _ ¸ ¸ _ ¸ ¸ _ 0 n ≥ 1, k≥0 (−1) k (2k+1)! n = 0, (−1) −n (−n)! k≥ −n−1 2 (−1) k (2k+n+1)! n ≤ −1 = _ ¸ ¸ ¸ ¸ _ ¸ ¸ ¸ ¸ _ 0 n ≥ 1, sin1 n = 0, 1 (−n)! cos 1 n ≤ −1, n is odd, 1 (−n)! sin1 n ≤ −1, n is even. Therefore sin z z + 1 = ∞ k=0 _ sin1 (2k)! (z + 1) −2k + cos 1 (2k + 1)! (z + 1) −(2k+1) _ . 4. (i) Proof. z = 0 is a removable singularity. (ii) Proof. z = 1 is a pole of order 1. The Laurent series of the function in a neighborhood of z = −1 can be obtained by 1 z 2 −1 cos πz z + 1 = 1 (z −1)(z + 1) ∞ n=0 (−1) n (2n)! _ πz z + 1 _ 2n . By discussion (3) of page 91, z = −1 is an essential singularity. 28 (iii) Proof. Since e 1 z = ∞ n=0 z −n , we have z(e 1 z −1) = ∞ n=1 z −n+1 . Therefore z = 0 is an essential singularity. (iv) Proof. z = 1 is an essential singularity. (v) Proof. z = 1 is an essential singularity. z = 0 is a pole of order 1. (vi) Proof. z = kπ + π 2 (k ∈ Z) are poles of order 1. 5. (1) Proof. The limit lim z→a f(z) exists (finite or ∞) if and only if lim z→a 1 f(z) exists. By discussion (3) on page 93, a is an essential singularity of f(z) if and only if a is an essential singularity of 1 f(z) . (2) Proof. Clearly, a is an isolated singularity of P(f(z)). We note by Fundamental Theorem of Algebra, P(ζ) maps C to C. Consequently, P(ζ) maps a dense subset of C to a dense subset of C. By Weierstrass Theorem (Theorem 3.3), f(z) maps a neighborhood of a to a dense subset of C. Therefore, P(f(z)) maps a neighborhood of a to a dense subset of C. So it is impossible for a to become a removable singularity or a pole of P(f(ζ)). Hence a must be also an essential singularity of P(f(z)). 6. (i) Proof. We note any positive integer can be uniquely represented in the form ∞ k=0 a k · 2 k , where each a k ∈ {0, 1} and only finitely many a k ’s are non-zero. In the expansion of ∞ n=0 (1+z 2 n ), each representation z ∞ k=0 a k ·2 k appears once and only once. Therefore, after a rearrangement of the terms, we must have ∞ n=0 (1 +z 2 n ) = 1 +z +z 2 +z 3 +z 4 +· · · = 1 1 −z . (ii) Proof. sinhπz = 1 2 [e πz − e −πz ] has zeros a n = ni (n ∈ Z), where a 0 = 0 is a zero of order 1. Since for any R > 0, we have ∞ n=−∞,n=0 _ R |a n | _ 2 = ∞ n=−∞,n=0 R 2 n 2 < ∞, by Weierstrass Factorization Theorem, we can find an entire function h(z) such that sinhπz = ze h(z) n=∞ n=−∞,n=0 __ 1 − z ni _ e z ni _ = ze h(z) ∞ n=1 _ 1 + z 2 n 2 _ . Since lim z→0 sinh πz z = π, we conclude e h(0) = π. Meanwhile we have π cothπz = (sinhπz) sinhπz = 1 sinh πz _ ze h(z) ∞ n=1 _ 1 + z 2 n 2 _ _ = 1 z +h (z) + ∞ n=1 2z n 2 1 + z 2 n 2 = 1 z +h (z) + ∞ n=1 2z n 2 +z 2 . 29 Let γ n be the rectangular path [n +(n + 1 2 )i, −n +(n + 1 2 )i, −n −(n + 1 2 )i, n −(n + 1 2 )i, n +(n + 1 2 )i]. Then for any given a, when n is large enough, we have _ γ n cothπz z 2 −a 2 dz = I +II +III +IV, where I = _ −n n e 2π[x+(n+ 1 2 )i] +1 e 2π[x+(n+ 1 2 )i] −1 _ x + (n + 1 2 )i ¸ 2 −a 2 dx = _ −n n −e 2πx +1 −e 2πx −1 _ x + (n + 1 2 )i ¸ 2 −a 2 dx, II = _ −(n+ 1 2 ) n+ 1 2 e 2π(−n+yi) +1 e 2π(−n+yi) −1 (−n +yi) 2 −a 2 idy, III = _ n −n e 2π[x−(n+ 1 2 )i] +1 e 2π[x−(n+ 1 2 )i] −1 [x −(n + 1 2 )i] 2 −a 2 dx = _ n −n −e 2πx +1 −e 2πx −1 [x −(n + 1 2 )i] 2 −a 2 dx and IV = _ n+ 1 2 −(n+ 1 2 ) e 2π(n+yi) +1 e 2π(n+yi) −1 (n +iy) 2 −a 2 idy. We note e x −1 e x +1 ∈ (−1, 1) for any x ∈ R. So we have |I| ≤ _ n −n 1 x 2 + (n + 1 2 ) 2 −a 2 dx = 2 _ (n + 1 2 ) 2 −a 2 arctan n _ (n + 1 2 ) 2 −a 2 → 0, as n → ∞. Similarly we can conclude lim n→∞ III = 0. We also note for x ∈ R large enough, e x +1 e x −1 < 2. So for n large enough, |II| ≤ _ n+ 1 2 −(n+ 1 2 ) e 1+e −2πn 1−e −2πn n 2 +y 2 −a 2 dy ≤ 4 √ n 2 −a 2 arctan n + 1 2 √ n 2 −a 2 → 0, as n → ∞. Similarly we can conclude lim n→∞ IV = 0. Combined, we have shown lim n→∞ _ γ n cothπz z 2 −a 2 dz = 0. Define f(z) = coth πz z 2 −a 2 . By Residue Theorem, for a ∈ iZ, we have _ γ n f(z)dz = 2πiRes(f; a) + 2πiRes(f; −a) + 2πi n k=−n Res(f; a n ). It’s easy to see Res(f; a) = Res(f; −a) = coth πa 2a . Since lim z→a n (z −a n )f(z) = lim z→a n e 2πz + 1 z 2 −a 2 z −a n e 2πz −1 = 2 2π(a 2 n −a 2 ) lim z→a n 2π(z −a n ) e 2πz −e 2πa n = − 1 π(n 2 +a 2 ) , we must have Res(f; a n ) = − 1 π(n 2 +a 2 ) . Therefore 0 = 1 2πi lim n→∞ _ γ n f(z)dz = lim n→∞ _ cothπa a − n k=−n 1 π(a 2 +k 2 ) _ = cothπa a − ∞ k=−∞ 1 π(a 2 +k 2 ) This implies for a ∈ iZ, π coth πa = 1 a + ∞ k=1 2a a 2 +n 2 . 30 Plugging this back into the formula π cothπz = 1 z + h (z) + ∞ n=1 2z n 2 +z 2 , we conclude h (z) = 0 for z ∈ iZ. By Theorem 2.13, h (z) ≡ 0 for any z ∈ C. So sinh πz = ze h(0) ∞ n=1 _ 1 + z 2 n 2 _ = πz ∞ n=1 _ 1 + z 2 n 2 _ . Remark 10. The above solution is a variant of the proof for factorization formula of sinπz. See Conway[2] VII, §6 for more details. (iii) Proof. From problem (v), we have π 2 ∞ n=1 _ 1 − 1 4n 2 _ = 1 and cos πz = sinπ _ 1 2 −z _ = π _ 1 2 −z _ ∞ n=1 _ 1 − _ 1 2 −z _ 2 n 2 _ = π _ 1 2 −z _ ∞ n=1 __ 1 + 1 2 −z n __ 1 − 1 2 −z n __ = π _ 1 2 −z _ ∞ n=1 __ 1 + 1 2n __ 1 − z n + 1 2 __ 1 − 1 2n __ 1 + z n − 1 2 __ = lim N→∞ π 2 _ 1 − z 1 2 _ N n=1 _ 1 − 1 4n 2 _ · N n=1 _ 1 − z n + 1 2 _ · N n=1 _ 1 + z n − 1 2 _ = lim N→∞ _ π 2 N n=1 _ 1 − 1 4n 2 _ _ · _ 1 − z N + 1 2 _ · N n=1 _ 1 − z 2 (n − 1 2 ) 2 _ = ∞ n=0 _ 1 − _ z n + 1 2 _ 2 _ . (iv) Proof. The function e πz − 1 has zeros a n = 2ni (n ∈ Z), where a 0 = 0 is a zero of order 1. Since for any R > 0, we have ∞ n=−∞,n=0 _ R |a n | _ 2 = ∞ n=−∞,n=0 R 2 4n 2 < ∞, by Weierstrass Factorization Theorem, we can find an entire function h(z) such that e πz −1 = ze h(z) ∞ n=−∞,n=0 __ 1 − z 2ni _ e z 2ni _ = ze h(z) ∞ n=1 _ 1 + z 2 4n 2 _ . To determine h(z), we note πe πz e πz −1 = (e πz −1) e πz −1 = 1 z +h (z) + ∞ n=1 2z 4n 2 1 + z 2 4n 2 = 1 z +h (z) + ∞ n=1 2z 4n 2 +z 2 . Let γ n be the rectangular path [2n+(2n+1)i, −2n+(2n+1)i, −2n−(2n+1)i, 2n−(2n+1)i, 2n+(2n+1)i]. Define f(z) = πe πz (z 2 −a 2 )(e πz −1) . Then for any given a, when n is large enough, we have _ γ n f(z)dz = I +II +III +IV, 31 where I = _ −2n 2n f(x + (2n + 1)i)dx = _ −2n 2n πe e πx [(x + (2n + 1)i) 2 −a 2 ](e πx + 1) dx, II = _ −(2n+1) 2n+1 f(−2n +yi)idy = _ −(2n+1) 2n+1 πe π(−2n+yi) [(−2n +yi) 2 −a 2 ](e π(−2n+yi) −1) idy, III = _ 2n −2n f(x −(2n + 1)i)dx = _ 2n −2n πe e πx [(x −(2n + 1)i) 2 −a 2 ](e πx + 1) dx, and IV = _ 2n+1 −(2n+1) f(2n +yi)idy = _ 2n+1 −(2n+1) πe π(2n+yi) [(2n +yi) 2 −a 2 ](e π(2n+yi) −1) idy. It is easy to see |I| ≤ _ 2n −2n πdx x 2 + (2n + 1) 2 −a 2 ≤ 2π _ (2n + 1) 2 −a 2 arctan 2n _ (2n + 1) 2 −a 2 → 0 as n → ∞. Similarly, we can show lim n→∞ III = 0. Also, we note |II| ≤ _ 2n+1 −(2n+1) πdy (4n 2 +y 2 −a 2 )(e 2nπ −1) ≤ 2π (e 2nπ −1) √ 4n 2 −a 2 arctan 2n + 1 √ 4n 2 −a 2 → 0, as n → ∞. For n sufficiently large, e 2nπ e 2nπ −1 ≤ 2, so |IV | ≤ _ 2n+1 −(2n+1) πe 2πn dy (4n 2 +y 2 −a 2 )(e 2πn −1) ≤ 4π √ 4n 2 −a 2 arctan 2n + 1 √ 4n 2 −a 2 → 0, as n → ∞. Combined, we have shown lim n→∞ _ γ n f(z)dz = 0. By Residue Theorem, for a ∈ 2Zi, we have _ γ n f(z)dz = 2πiRes(f; a) + 2πiRes(f; −a) + 2πi n k=−n Res(f; a n ). It’s easy to see Res(f; a) = πe πa 2a(e πa −1) and Res(f; −a) = π 2a(e πa −1) . Since lim z→a n (z −a n )f(z) = lim z→a n πe πz (z 2 −a 2 ) · z −a n e πz −1 = − 1 4n 2 +a 2 , Res(f; a n ) = − 1 4n 2 +a 2 . Therefore 0 = 1 2πi lim n→∞ _ γ n f(z)dz = lim n→∞ _ π 2a e πa + 1 e πa −1 − n k=−n 1 4n 2 +a 2 _ . So π 2a e πa +1 e πa −1 = 1 a 2 + ∞ n=1 2 4n 2 +a 2 for a ∈ 2Zi. Therefore for z ∈ 2Zi, we have h (z) = πe πz e πz −1 − 1 z − ∞ n=1 2z 4n 2 +z 2 = πe πz e πz −1 − π 2 e πz + 1 e πz −1 = π 2 . So h(z) = πz 2 +h(0). It’s easy to see lim z→0 e πz −1 z = π. So e h(0) = π. Combined, we conclude e h(z) = πe πz 2 . So e πz −1 = πze πz 2 ∞ n=1 _ 1 + z 2 4n 2 _ . 32 Equivalently, we have e z −1 = ze z 2 ∞ n=1 _ 1 + z 2 4π 2 n 2 _ . (v) Proof. Using the formula sinπz = −i sinh(iz) and the factorization formula for sinhπz, we have sinπz = −iπ(iz) ∞ n=1 _ 1 − z 2 n 2 _ = πz ∞ n=1 _ 1 − z 2 n 2 _ . 7. For a clear presentation of convergence of infinite products, we refer to Conway [2], Chapter VII, §5. In particular, we quote the following theorem (Conway [2], Chapter VII, §5, Theorem 5.9) Let G be an open subset of the complex plane. Denote by H(G) the collection of analytic functions on G, equipped with the topology determined by uniform convergence on compact subsets of G. Then H(G) is a complete metric space. Furthermore, we have the following theorem (proof omitted). Theorem 11. Let G be a region in C and let (f n ) ∞ n=1 be a sequence in H(G) such that no f n is identically zero. If [f n (z)−1] converges absolutely and uniformly on compact subsets of G, then ∞ n=1 f n (z) converges in H(G) to an analytic function f(z). If a is a zero of f then a is a zero of only a finite number of the functions f n , and the multiplicity of the zero of f at a is the sum of the multiplicities of the zeros of the functions f n at a. Proof. (Proof of Blaschke Product) Fix r ∈ (0, 1). Since ∞ k=1 (1 − |a k |) < ∞, lim k→∞ |a k | = 1. So there exists k 0 ∈ N, such that for any k ≥ k 0 , 1+r 2 < |a k | < 1. So for any k ≥ k 0 , ¸ ¸ ¸ ¸ a k −z 1 − ¯ a k z · |a k | a k −1 ¸ ¸ ¸ ¸ ≤ |a k |(1 −|a k |) +|z||a k |(1 −|a k |) |1 − ¯ a k z||a k | ≤ (1 +r)(1 −|a k |) 1+r 2 (1 −r) = 2 1 −r (1 −|a k |). Since ∞ k=1 (1 − |a k |) < ∞, we conclude ∞ k=1 ¸ ¸ ¸ a k −z 1−¯ a k z · |a k | a k −1 ¸ ¸ ¸ is absolutely and uniformly convergent on {z : |z| ≤ r}. Therefore ∞ k=1 a k −z 1−¯ a k z · |a k | a k is absolutely and uniformly convergent on {z : |z| ≤ r}. By Weierstrass Theorem (Theorem 3.1), f(z) = ∞ k=1 a k −z 1−¯ a k z · |a k | a k represents a non-zero holomorphic function on {z : |z| < 1}. Since the mapping z → a k −z 1−¯ a k z (0 < |a k | < 1) maps D(0, 1) to D(0, 1), we conclude |f(z)| ≤ 1. By the theorem quoted at the beginning of the solution, it’s clear that (a k ) ∞ k=1 are the only zeros of f(z). 8. Proof. Let R ε = R −ε, ε ∈ (0, R). Define F ε (z) = f(z) · t j=1 R ε (z−b j ) R 2 ε − ¯ b j z s i=1 R ε (z−a i ) R 2 ε −¯ a i z , ∀z ∈ D(0, R ε ). It’s easy to verify that each of R 2 ε (z−b j ) R 2 ε − ¯ b j z and R 2 ε (z−a i ) R 2 ε −¯ a i z maps D(0, R ε ) onto itself and takes the boundary to the boundary. Therefore F ε (z) is analytic on D(0, R ε ), has no zeros in D(0, R ε ), and |F ε (z)| = |f(z)| for |z| = R ε . Since log |F ε (z)| is a harmonic function, by formula (2.33) on page 71, log |F ε (z)| = 1 2π _ 2π 0 log |F ε (R ε e iφ )|Re R ε e iφ +z R ε e iφ −z dφ = 1 2π _ 2π 0 log |f(R ε e iφ )|Re R ε e iφ +z R ε e iφ −z dφ. 33 Plugging the formula of F ε into the above equality, we get log |f(z)| = 1 2π _ 2π 0 log |f(R ε e iφ )|Re R ε e iφ +z R ε e iφ −z dφ + t j=1 log ¸ ¸ ¸ ¸ R 2 ε − ¯ b j z R ε (z −b j ) ¸ ¸ ¸ ¸ − s i=1 log ¸ ¸ ¸ ¸ R 2 ε − ¯ a i z R ε (z −a i ) ¸ ¸ ¸ ¸ . Letting ε → 0 yields the desired formula. 9. Proof. By Theorem 3.4 (correction: “holomorphic function f(z)” in the theorem’s statement should be “meromorphic function f(z)”), f(z) has the form of 1 z+1 + 2 z−2 + 3 (z−2) 2 +c, where c is a constant. f(0) = 7 4 implies c = 1. Since for any ζ ∈ D(0, 1), 1 (1 −ζ) 2 = _ 1 1 −ζ _ = _ ∞ n=0 ζ n _ = ∞ n=0 (n + 1)ζ n , the Laurent expansion of f(z) in 1 < |z| < 2 is f(z) = 1 z + 1 + 2 z −2 + 3 (z −2) 2 + 1 = 1 z 1 1 + 1 z − 1 1 − z 2 + 3 4 · 1 (1 − z 2 ) 2 + 1 = 1 z ∞ n=0 _ − 1 z _ n − ∞ n=0 _ z 2 _ n + 3 4 ∞ n=0 (n + 1) _ z 2 _ n + 1 = − ∞ n=1 (−1) n z n + 3 4 + ∞ n=1 3n −1 2 n+2 z n . 10. Proof. We follow the construction outlined in the proof of Mittag-Leffler Theorem (Theorem 3.9). For each n ∈ N, when |z| < n 2 , ψ n (z) has Taylor expansion ψ n (z) = n (z −n) 2 = 1 n · 1 _ 1 − z n _ 2 = 1 n ∞ k=0 (k + 1) _ z n _ k . Let λ n be a positive integer to be determined later. We estimate the tail error obtained by retaining only the first λ n terms of the Taylor expansion of ψ n (z): ¸ ¸ ¸ ¸ ¸ ψ n (z) − 1 n λ n k=0 (k + 1) _ z n _ k ¸ ¸ ¸ ¸ ¸ ≤ 1 n ∞ k=λ n +1 (k + 1) 2 k = 1 n ∞ k=λ n (k + 2) 2 k+1 = 1 n _ ∞ k=λ n k 2 k+1 + 1 2 λ n −1 _ ≤ 1 n · 2 λ n −1 + 1 n ∞ k=λ n _ k+1 k x 2 x dx = 1 n · 2 λ n −1 + 1 n _ ∞ λ n x log 1 2 d _ 1 2 _ x = 1 n · 2 λ n −1 + 1 n · 2 λ n _ λ n log 2 + 1 (log 2) 2 _ . If we let λ n = n, then ε n := 1 n·2 λ n −1 + 1 n·2 λ n _ λ n log 2 + 1 (log 2) 2 _ = 1 n·2 n−1 + 1 n·2 n _ n log 2 + 1 (log 2) 2 _ satisfies ∞ n=1 ε n < ∞. By the proof of Mittag-Leffler Theorem, we can write f(z) as f(z) = U(z) + ∞ n=1 _ n (z −n) 2 − 1 n n k=0 (k + 1) _ z n _ k _ , where U(z) is an entire function. 34 11. (i) Proof. Define f(z) = 1 (z 2 −a 2 )(e πz −1) , where a ∈ C\2iZ. Let γ n be the rectangular path [2n+(2n+1)i, −2n+ (2n +1)i, −2n −(2n +1)i, 2n −(2n +1)i, 2n +(2n +1)i]. Then for any given a, when n is large enough, we have _ γ n f(z)dz = I +II +III +IV, where I = _ −2n 2n f(x + (2n + 1)i)dx = _ 2n −2n dx [(x + (2n + 1)i) 2 −a 2 ](e πx + 1) , II = _ −(2n+1) 2n+1 f(2n +yi)idy = _ −(2n+1) 2n+1 idy [(−2n +yi) 2 −a 2 ](e π(−2n+yi −1) , III = _ 2n −2n f(x −(2n + 1)i)dx = _ 2n −2n −dx [(x −(2n + 1)i) 2 −a 2 ](e πx + 1) , and IV = _ 2n+1 −(2n+1) f(2n +iy)idy = _ 2n+1 −(2n+1) idy [(2n +iy) 2 −a 2 ](e π(2n+iy) −1) . It’s easy to see |I| ≤ _ 2n −2n dx x 2 + (2n + 1) 2 −a 2 ≤ 2 _ (2n + 1) 2 −a 2 arctan 2n _ (2n + 1) 2 −a 2 → 0 as n → ∞. Similarly, we can show lim n→∞ III = 0. Also, we note |II| ≤ _ 2n+1 −(2n+1) dy (4n 2 +y 2 −a 2 )(1 −e −2nπ ) = 2 (1 −e −2nπ ) √ 4n 2 −a 2 arctan 2n + 1 √ 4n 2 −a 2 → 0 as n → ∞, and |IV | ≤ _ 2n+1 −(2n+1) dy (4n 2 +y 2 −a 2 )(e 2πn −1) = 2 (e 2πn −1) √ 4n 2 −a 2 arctan 2n + 1 4n 2 −a 2 → 0 as n → ∞. Combined, we have shown lim n→∞ _ γ n f(z)dz = 0. By Residue Theorem, for a ∈ 2iZ, we have _ γ n f(z)dz = 2πiRes(f; a) + 2πiRes(f; −a) + 2πi n k=−n Res(f; 2ni). It’s easy to see Res(f; a) = 1 2a(e πa −1) and Res(f; −a) = 1 2a(1−e −πa ) . Since lim z→2ni (z −2ni)f(z) = lim z→2ni π(z −2ni) π(z 2 −a 2 )(e πz −1) = − 1 π(4n 2 +a 2 ) , we have Res(f; 2ni) = − 1 π(4n 2 +a 2 ) . Therefore 0 = 1 2πi lim n→∞ _ γ n f(z)dz = lim n→∞ _ 1 2a · e πa + 1 e πa −1 − 1 π n k=−n 1 4k 2 +a 2 _ = 1 2a · e πa + 1 e πa −1 − 1 π ∞ k=−∞ 1 4k 2 +a 2 . Replacing πa with z, we get after simplification 1 e z −1 = 1 z − 1 2 + ∞ k=1 2z 4k 2 π 2 +z 2 , ∀z ∈ C \ 2Zπi. This is the partial fraction of 1/(e z −1). 35 (ii) Proof. Define f(z) = 1 (z−a) sin 2 (πz) , where a ∈ C\Z. Let γ n be the rectangular path [ _ n + 1 2 _ +ni, − _ n + 1 2 _ + ni, − _ n + 1 2 _ −ni, _ n + 1 2 _ −ni, _ n + 1 2 _ +ni]. Then when n is large enough, we have _ γ n f(z)dz = I +II +III +IV, where I = _ −(n+ 1 2 ) n+ 1 2 dx (x +ni −a) sin 2 [π(x +ni)] = _ (n+ 1 2 ) −(n+ 1 2 ) 4dx (x +ni −a)(e −2nπ+2ixπ +e 2nπ−2ixπ −2) , II = _ −n n idy [−(n + 1 2 ) +yi −a] sin 2 [−π(n + 1 2 ) +πyi] = _ n −n −4idy [−(n + 1 2 ) −a +yi](e 2πy +e −2πy + 2) , III = _ (n+ 1 2 ) −(n+ 1 2 ) dx (x −ni −a) sin 2 [π(x −ni)] = _ (n+ 1 2 ) −(n+ 1 2 ) −4dx (x −a −ni)(e 2nπ+2iπx +e −2nπ−2iπx −2) , and IV = _ n −n idy [(n + 1 2 ) +yi −a] sin 2 [π(n + 1 2 ) +πyi] = _ n −n 4idy [(n + 1 2 ) −a +yi](e −2yπ +e 2yπ + 2) . It’s easy to see |I| ≤ _ (n+ 1 2 ) −(n+ 1 2 ) 4dx _ (x −a) 2 +n 2 (e 2nπ −e −2nπ −2) ≤ 4(2n + 1) n(e 2nπ −e −2nπ −2) → 0 as n → ∞, |II| ≤ _ n −n 4dy _ (n + 1 2 +a) 2 +y 2 (e 2πy +e −2πy + 2) = _ n 0 8dy _ (n + 1 2 +a) 2 +y 2 (e 2πy +e −2πy + 2) ≤ _ n 0 8dy (n + 1 2 +a)e 2πy = 8 (n + 1 2 +a) 1 −e −2πn 2π → 0 as n → ∞, |III| ≤ _ (n+ 1 2 ) −(n+ 1 2 ) 4dx _ (x −a) 2 +n 2 (e 2nπ −e −2nπ −2) ≤ 4(2n + 1) n(e 2nπ −e −2nπ −2) → 0 as n → ∞, and lastly |IV | ≤ _ n −n 4dy _ (n + 1 2 −a) 2 +y 2 (e 2yπ +e −2yπ + 2) = _ n 0 8dy _ (n + 1 2 −a) 2 +y 2 (e 2yπ +e −2yπ + 2) ≤ _ n 0 8 (n + 1 2 −a)e 2yπ = 8 (n + 1 2 −a) 1 −e −2πn 2π → 0 as n → ∞. Combined, we have shown lim n→∞ _ γ n f(z)dz = 0. By Residue Theorem, we have _ γ n f(z)dz = 2πiRes(f; a) + 2πi n k=−n Res(f; k). 36 It’s easy to see Res(f; a) = 1 sin 2 (πa) . Define h k (z) = (z−k) 2 sin 2 (πz) . Since lim z→k h k (z) = 1 π 2 , h k is holomorphic in a neighborhood of k and for ε > 0 small enough, and we have Res(f; k) = 1 2πi _ |z−k|=ε dz (z −a) sin 2 (πz) = 1 2πi _ |z−k|=ε h k (z) (z −a) · dz (z −k) 2 = d dz _ h k (z) z −a _¸ ¸ ¸ ¸ z=k . We note d dz _ h k (z) z −a _¸ ¸ ¸ ¸ z=k = lim z→k h k (z)(z −a) −h(z) (z −a) 2 = lim z→k h k (z) z −a − 1 (k −a) 2 π 2 , and lim z→k h k (z) z −a = lim z→k 2(z −k) sin(πz) −2π(z −k) 2 cos(πz) sin 3 (πz) = lim z→k 2 sin(πz) + 2π(z −k) cos(πz) −4π(z −k) cos(πz) + 2π 2 (z −k) 2 sin(πz) 3 sin 2 (πz) · π cos(πz) = lim z→k 2 sin(πz) −2π(z −k) cos(πz) 3π sin 2 (πz) = lim z→k 2π cos(πz) −2π cos(πz) + 2π 2 (z −k) sin(πz) 6π 2 sin(πz) cos(πz) = 0. So for a ∈ C \ Z, _ γ n f(z)dz = 2πi sin 2 (πa) − n k=−n 2πi (k −a) 2 π 2 . Let n → ∞, we get π 2 sin 2 (πz) = ∞ n=−∞ 1 (z −n) 2 . Remark 12. Another choice is to let f(z) = cot(πz) (z+a) 2 , see Conway [2], page 122, Exercise 6. The trick used in this problem and problem 6 (ii) will be generalized in problem 12. (iii) Proof. By the solution of problem 6 (ii), we have coth z = 1 z + ∞ n=1 2z z 2 +π 2 n 2 . Since cothz = i cot(iz), we have cot z = i coth(iz) = 1 z − ∞ n=1 2z π 2 n 2 −z 2 = 1 z − ∞ n=1 _ 1 πn −z − 1 πn +z _ . Therefore π α cot πβ α = 1 β − ∞ n=1 _ 1 αn −β − 1 αn +β _ = lim N→∞ _ 1 β − N n=1 _ 1 αn −β − 1 αn +β _ _ = lim N→∞ _ 1 β − _ 1 α −β + 1 2α −β +· · · + 1 Nα −β _ + _ 1 α +β + 1 2α +β +· · · + 1 Nα +β __ = lim N→∞ __ 1 β − 1 α −β _ + _ 1 α +β − 1 2α −β _ +· · · + _ 1 (N −1)α +β − 1 Nα −β _ + 1 Nα +β _ = ∞ n=0 _ 1 nα +β − 1 nα + (α −β) _ = ∞ n=0 α −2β (nα +β)[nα + (α −β)] . 37 By letting α = 3 and β = 1, we get π 3 √ 3 = π 3 cot π 3 = ∞ n=1 1 (3n −2)(3n −1) . Remark 13. For a direct proof without using problem 6 (ii), we can let f(z) = cot(πz) z 2 −a 2 and apply Residue Theorem to it. For details, see Conway [2], page 122, Exercise 8. The line of reasoning used in this approach will be generalized in problem 12 below. 12. (1) Proof. Since z = ∞ is a zero of f(z) with multiplicity p, f(z) can be written as h(z) z p where h is holomorphic in a neighborhood of ∞. Consequently, h is bounded in a neighborhood of ∞. In the below, we shall work with a sequence of neighborhoods of ∞ that shrinks to ∞. So without loss of generality, we can assume h is bounded and denote its bounded by M. Let γ n be the rectangular path [(n + 1 2 ) + ni, −(n + 1 2 ) + ni, −(n + 1 2 ) − ni, (n + 1 2 ) − ni, (n + 1 2 ) + ni]. Then _ γ n f(z) cot(πz)dz = I +II +III +IV, where I = _ −(n+ 1 2 ) (n+ 1 2 ) h(x +ni) (x +ni) p e iπ(x+ni) +e −iπ(x+ni) 2 e iπ(x+ni) −e −iπ(x+ni) 2i dx = _ −(n+ 1 2 ) (n+ 1 2 ) h(x +ni) (x +ni) p e iπx−nπ +e −iπx+nπ e iπx−nπ −e −iπx+nπ idx, II = _ −n n h(−(n + 1 2 ) +yi) [−(n + 1 2 ) +yi] p e iπ[−(n+ 1 2 )+yi] +e −iπ[−(n+ 1 2 )+yi] 2 e iπ[−(n+ 1 2 )+yi] −e −iπ[−(n+ 1 2 )+yi] 2i idy = _ −n n h(−(n + 1 2 ) +yi) [−(n + 1 2 ) +yi] p (−1) n e −yπ (−i) + (−1) n e yπ i (−1) n e −yπ (−i) −(−1) n e yπ i (−1)dy, III = _ (n+ 1 2 ) −(n+ 1 2 ) h(x −ni) (x −ni) p e iπx+nπ +e −iπx−nπ e iπx+nπ −e −iπx−nπ idx, and IV = _ n −n h((n + 1 2 ) +yi) [(n + 1 2 ) +yi] p (−1) n e −yπ i + (−1) n e yπ (−i) (−1) n e −yπ i −(−1) n e yπ (−i) (−1)dy. We note |I| ≤ _ (n+ 1 2 ) −(n+ 1 2 ) M (x 2 +n 2 ) p 2 e −nπ +e nπ e nπ −e −nπ dx = e 2nπ + 1 e 2nπ −1 2M n arctan n + 1 2 n → 0 as n → ∞, |II| ≤ _ n −n M _ (n + 1 2 ) 2 +y 2 ¸ p 2 ¸ ¸ ¸ ¸ e yπ −e −yπ e yπ +e −yπ ¸ ¸ ¸ ¸ dy ≤ 2M n + 1 2 arctan n n + 1 2 → 0 as n → ∞, |III| ≤ _ (n+ 1 2 ) −(n+ 1 2 ) M (x 2 +n 2 ) p 2 e nπ +e −nπ e nπ −e −nπ dx = e nπ +e −nπ e nπ −e −nπ 2M n arctan n + 1 2 n → 0 as n → ∞, and |IV | ≤ _ n −n M _ (n + 1 2 ) 2 +y 2 ¸ p 2 ¸ ¸ ¸ ¸ e yπ −e −yπ e yπ +e −yπ ¸ ¸ ¸ ¸ dy ≤ 2M n + 1 2 arctan n n + 1 2 → 0 38 as n → ∞. Combined, we can conclude lim n→∞ _ γ n f(z) cot(πz)dz = 0. Meanwhile, for n sufficiently large, by Residue Theorem, we have _ γ n f(z) cot(πz)dz = 2πi m k=1 Res(f(z) cot(πz), α k ) + 2πi n k=−n Res(f(z) cot(πz), k). We note Res(f(z) cot(πz), k) = 1 2πi _ |z−k|=ε f(z) cos(πz) sin(πz) dz = 1 2πi _ |z−k|=ε (−1) k f(z) z −k · cos(πz)(z −k) sin[π(z −k)] dz = f(k) π . So _ γ n f(z) cot(πz)dz = 2πi m k=1 Res(f(z) cot(πz), α k ) + 2πi n k=−n f(k) π . Letting n → ∞ gives us lim n→∞ n k=−n f(k) = −π n k=1 Res(f(z) cot(πz), α k ). (2) Proof. As argued in the solution of (1), we can assume f(z) = h(z) z p where h is holomorphic in a neighborhood of ∞ and has bound M in that neighborhood. Let γ n denote the same rectangular path as in our solution of (1). Then _ γ n f(z) sin(πz) dz = I +II +III +IV, where I = _ −(n+ 1 2 ) n+ 1 2 h(x +ni) (x +ni) p 2idx e iπx−nπ −e −iπx+nπ , II = _ −n n h(−(n + 1 2 ) +yi) [−(n + 1 2 ) +yi] p 2i e −yπ (−1) n (−i) −e yπ (−1) n i idy, III = _ n+ 1 2 −(n+ 1 2 ) h(x −ni) (x −ni) p 2idx e iπx+nπ −e −iπx−nπ , and IV = _ n −n h((n + 1 2 ) +yi) _ (n + 1 2 ) +yi ¸ p 2i e −πy (−1) n i −e πy (−1) n (−i) idy. We note |I| ≤ 2M e nπ −e −nπ _ n+ 1 2 −(n+ 1 2 ) dx (x 2 +n 2 ) p 2 ≤ 4M e nπ −e −nπ _ n+ 1 2 0 dx x 2 +n 2 = 2M (e nπ −e −nπ )n arctan n + 1 2 n → 0 as n → ∞, and |II| ≤ _ n −n 1 _ (n + 1 2 ) 2 +y 2 ¸ p 2 2Mdy e yπ +e −yπ ≤ _ n −n Mdy (n + 1 2 ) 2 +y 2 = 2M n + 1 2 arctan n n + 1 2 → 0 as n → ∞. By similar argument, we can prove lim n→∞ III = lim n→∞ IV = 0. Combined, we can conclude that lim n→∞ _ γ n f(z) sin(πz) dz = 0. Meanwhile, for n sufficiently large, by Residue Theorem, we have _ γ n f(z) sin(πz) dz = 2πi m k=1 Res _ f(z) sin(πz) , α k _ + 2πi n k=−n Res _ f(z) sin(πz) , k _ . 39 It’s easy to see Res _ f(z) sin(πz) , k _ = 1 2πi _ |z−k|=ε (−1) k z −k · f(z)(z −k) sin[π(z −k)] dz = (−1) k f(k) π . Therefore, _ γ n f(z) sin(πz) dz = 2πi m k=1 Res _ f(z) sin(πz) , α k _ + 2πi n k=−n (−1) k f(k) π . Let n → ∞, we get lim n→∞ n k=−n (−1) k f(k) = −π m k=1 Res _ f(z) sin(πz) , α k _ . (3) (i) Proof. Let f(z) = 1 (z+a) 2 . Then by result of (1), ∞ n=−∞ 1 (n +a) 2 = −πRes _ cot(πz) (z +a) 2 , −a _ = −π d dz [cot(πz)]| z=−a = π 2 sin 2 (πa) . This result is verified by problem 11 (ii). (ii) Proof. Let f(z) = 1 z 2 +a 2 . Then by result of (2), ∞ n=−∞ (−1) n n 2 +a 2 = −π _ Res _ 1 (z 2 +a 2 ) sin(πz) , ai _ + Res _ 1 (z 2 +a 2 ) sin(πz) , −ai __ = − π ai · sin(πai) = πcsch(πa) a . This result can be verified by Mathematica. 13. (i) Proof. f(z) = 1 z 2 −z 4 has poles 0, 1, and −1. ∞ is a removable singularity. Res(f; 0) = 1 2πi _ |z|=ρ dz z 2 (1 −z 2 ) = d dz _ 1 1 −z 2 _ | z=0 = −(−2z) (1 −z 2 ) 2 | z=0 = 0, Res(f; 1) = 1 2πi _ |z−1|=ρ dz (1 −z)(1 +z)z 2 = − 1 (1 +z)z 2 | z=1 = − 1 2 , and Res(f; −1) = 1 2πi _ |z+1|=ρ dz (1 +z)(1 −z)z 2 = 1 2 . (ii) Proof. We note f(z) := z 2 +z + 2 z(z 2 + 1) 2 = 2 z + −1 +i 4(z −i) 2 + −4 −i 4(z −i) + −1 −i 4(z +i) 2 + −4 +i 4(z +i) . Therefore, the function f(z) has poles 0, i and −i. ∞ is a removable singularity of f(z). Furthermore, we have Res(f; 0) = 2, Res(f; i) = −4 −i 4 = −1 − i 4 , Res(f; −i) = −4 +i 4 = −1 + i 4 . 40 (iii) Proof. Suppose a 1 , · · · , a n are the roots of the equation z n + a n = 0. Then each a i is a pole of order 1 for the function f(z) = z n−1 z n +a n . ∞ is a removable singularity of f(z). Then Res(f; a i ) = a n−1 i n j=1,j=i (a i −a j ) = lim z→a i a n−1 i (z −a i ) n j=1 (z −a j ) = lim z→a i a n−1 i (z −a i ) z n +a n = 1 n . (iv) Proof. f(z) = 1 sin z has πZ as poles of order 1. ∞ is not an isolated singularity of f(z). And Res(f, nπ) = 1 2πi _ |z−nπ|=ε dz sinz = 1 2πi _ |z−nπ|=ε 1 z −nπ (−1) n (z −nπ) sin(z −nπ) dz = (−1) n . (v) Proof. First, we note as cos _ 1 z−2 _ = 1 2 _ e i z−2 +e − i z−2 _ = ∞ n=0 (−1) n (z−2) 2n . So f(z) := z 3 cos _ 1 z−2 _ = ∞ n=0 (−1) n z 3 (z−2) 2n . This shows z = 2 is an essential singularity of f(z), and Res(f; 2) = ∞ n=0 (−1) n 2πi _ |z−2|=ε z 3 (z −2) 2n dz = (−1) · 3z 2 | z=2 + (−1) 2 · 1 3! · 6| z=2 = −11. Also, ∞ is an isolated singularity of f(z), and we have Res(f; ∞) = − 1 2πi _ |z|=R z 3 cos _ 1 z −2 _ dz = − 1 2πi _ |z−2|=R z 3 cos _ 1 z −2 _ dz = − 1 2πi _ |ζ|= 1 R _ 1 ζ + 2 _ 3 cos ζ _ − dζ ζ 2 _ = 1 2πi _ |ζ=ε| (ζ + 2) 3 ζ 5 cos ζdζ = 1 4! d 4 dζ 4 _ (ζ + 2) 3 cos ζ ¸ | ζ=0 = − 8 3 . (vi) Proof. Let f(z) = e z z(z+1) = e z _ 1 z − 1 z+1 _ . Then Res(f; 0) = 1, Res(f; −1) = −e −1 , and Res(f; ∞) = 1 −e −1 . 14. Proof. We can write g(z) as (z −a) 2 h(z) where h is holomorphic and h(a) = 0. Then h(a) = g(a) (z−a) 2 and h (a) = g (a)(z −a) −2g(a) (z −a) 3 . 41 Therefore Res _ f(z) g(z) , a _ = Res _ 1 (z −a) 2 · f(z) h(z) , a _ = d dz _ f(z) h(z) _¸ ¸ ¸ ¸ z=a = f (a)h(a) −f(a)h (a) h 2 (a) = f (a) g(a) (z−a) 2 −f(a) g (a)(z−a)−2g(a) (z−a) 3 g 2 (a) (z−a) 4 = g(a)(z −a)[f (a)(z −a) + 2f(a)] −f(a)g (a)(z −a) 2 g 2 (a) . 15. We make an observation of some simple rules that facilitate the evaluation of integrals via Residue Theorem. Rule 1 (rule for integrand function). The poles of the integrand function should be easy to find, such that the integrand function can be easily represented as f(z) (z−a) n , where f is holomorphic. The holomorphic function f(z) can be multi-valued function like logarithm function and power function. When it’s difficult to make f holomorphic in the desired region, check if it is the real or imaginary part of a holomorphic function. Rule 2 (rule for integration path). The choice of integration path is highly dependent on the properties of the integrand function, where symmetry and multi-valued functions (log and power functions) are often helpful. Oftentimes, the upper and lower limits of integration either form a full circle (i.e. [0, 2π]) so that substitution for trigonometric functions can be easily done or have ∞ as one of the end points. (i) Proof. Let γ 1 = {z : −R ≤ Rez ≤ R, Imz = 0}, γ 2 = {z : |z| = R, 0 ≤ arg z ≤ π}, and f(z) = z 2 (z 2 +1) 2 . Then _ γ 1 f(z)dz + _ γ 2 f(z)dz = 2πi · Res(f, i) = 2πi · lim z→i d dz _ z 2 (z +i) 2 _ = π 2 , and ¸ ¸ ¸ ¸ _ γ 2 f(z)dz ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ _ π 0 R 2 e 2θi (R 2 e 2θi + 1) 2 Re iθ · idθ ¸ ¸ ¸ ¸ ≤ _ π 0 R 3 (R 2 −1) 2 dθ → 0 as R → ∞. So by letting R → ∞, we have _ ∞ 0 x 2 dx (x 2 +1) 2 = 1 2 _ ∞ −∞ x 2 dx (x 2 +1) 2 = π 4 . (ii) Proof. We note sin 2 x = 1−cos 2x 2 , so _ π 2 0 dx a + sin 2 x = 1 2 _ π 0 dx a + sin 2 x = 1 2 _ 2π 0 dθ (2a + 1) −cos θ . Let b = 2a + 1, then _ π 2 0 dx a + sin 2 x = 1 2 _ |z|=1 dz iz b − 1 2 (z +z −1 ) = 1 2i _ |z|=1 dz − 1 2 z 2 +bz − 1 2 . The equation − 1 2 z 2 + bz − 1 2 = 0 has two roots: z 1 = b − √ b 2 −1 and z 2 = b + √ b 2 −1. Since b > 1, we have z 1 ∈ D(0, 1) and z 2 ∈ ¯ D(0, 1). Therefore, _ π 2 0 dx a + sin 2 x = i _ |z|=1 dz (z −z 1 )(z −z 2 ) = i · 2πi · 1 z 1 −z 2 = π √ b 2 −1 = π 2 _ a(a + 1) . 42 Remark 14. If R(x, y) is a rational function of two variables x and y, for z = e iθ , we have R(sinθ, cos θ) = R _ 1 2 _ z + 1 z _ , 1 2i _ z − 1 z __ , dθ = dz iz . Therefore _ 2π 0 R(sinθ, cos θ)dθ = _ |z|=1 R _ 1 2 (z + 1 z ), 1 2i (z − 1 z ) _ dz iz . Remark 15. When the integrand function is a rational function of trigonometric functions, in view of the previous remark, it is desirable to have [0, 2π] as the integration interval. For this reason, we first use symmetry to expand the integration interval from [0, π 2 ] to [0, π], and then use double angle formula to expand [0, π] to [0, 2π]. This solution is motivated by Rule 2. (iii) Proof. Let γ 1 = {z : −R ≤ Rez ≤ R, Imz = 0}, γ 2 = {z : |z| = R, 0 ≤ arg z ≤ π}, and f(z) = ze iz z 2 +1 . Then _ γ 1 f(z)dz + _ γ 2 f(z)dz = 2πiRes(f, i) = _ |z−i|=ε ze iz dz (z −i)(z +i) = 2πi · ie −1 2i = π e i, and ¸ ¸ ¸ ¸ _ γ 2 f(z)dz ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ ¸ _ π 0 Re iRe iθ R 2 e 2θi + 1 Re iθ · idθ ¸ ¸ ¸ ¸ ¸ ≤ _ π 0 R 2 e −Rsin θ R 2 −1 dθ → 0 as R → ∞ by Lebesgue’s Dominated Convergence Theorem. Therefore by letting R → ∞, we have _ ∞ −∞ xe ix x 2 + 1 dx = _ ∞ −∞ xcos x x 2 + 1 dx +i _ ∞ −∞ xsinx x 2 + 1 dx = π e i, which implies _ ∞ 0 xsin x x 2 +1 dx = π 2e . (iv) Proof. Let r, R be two positive numbers such that r < 1 < R. Let γ 1 = {z : r ≤ |z| ≤ R, arg z = 0}, γ 2 = {z : r ≤ |z| ≤ R, arg z = π}, γ R = {z : |z| = R, 0 ≤ arg z ≤ π}, and γ r = {z : |z| = r, 0 ≤ arg z ≤ π}. Define f(z) = log z (1+z 2 ) 2 where log z is defined on C \ [0, ∞) and takes the principle branch of Logz with arg z ∈ (0, 2π) (see page 23). By Residue Theorem, _ γ 1 +γ R +γ 2 −γ r f(z)dz = 2πiRes(f, i) = 2πi lim z→i d dz _ log z (z +i) 2 _ = − π 2 + π 2 4 i. We note ¸ ¸ ¸ ¸ _ γ R f(z)dz ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ _ π 0 log(Re iθ ) (R 2 e 2iθ + 1) 2 Re iθ · idθ ¸ ¸ ¸ ¸ ≤ _ π 0 R _ (log R) 2 +θ 2 (R 2 −1) 2 dθ ≤ πR(log R +π) (R 2 −1) 2 → 0 as R → ∞, ¸ ¸ ¸ ¸ _ γ r f(z)dz ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ _ π 0 log(re iθ ) (r 2 e 2iθ + 1) 2 re iθ · idθ ¸ ¸ ¸ ¸ ≤ _ π 0 r _ (log r) 2 +θ 2 (1 −r 2 ) 2 dθ ≤ πr(−log r +π) (1 −r 2 ) 2 → 0 as r → 0, and _ γ 2 f(z)dz = _ −r −R log x (1 +x 2 ) 2 dx = _ R r log(−x) (1 +x 2 ) 2 dx = _ R r log x +πi (1 +x 2 ) 2 dx. 43 Therefore by letting r → 0 and R → ∞, we have _ ∞ 0 2 log x +πi (1 +x 2 ) 2 dx = − π 2 + π 2 4 i, i.e. _ ∞ 0 log x (1+x 2 ) 2 dx = − π 4 . (v) Proof. Let γ 1 , γ 2 , γ r and γ R be defined as in (iv). Define f(z) = z 1−α 1+z 2 , where z 1−α = e (1−α) log z is defined on C \ [0, ∞) with log z taking the principle branch of Logz. By Residue Theorem, _ γ 1 +γ R +γ 2 −γ r f(z)dz = 2πiRes(f, i) = 2πi · e (1−α) log z z +i | z=i = πe (1−α) π 2 i . We note ¸ ¸ ¸ ¸ _ γ R f(z)dz ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ _ π 0 R 1−α e iθ(1−α) 1 +R 2 e 2iθ · Re iθ · idθ ¸ ¸ ¸ ¸ ≤ R 2−α R 2 −1 π → 0 as R → ∞, ¸ ¸ ¸ ¸ _ γ r f(z)dz ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ _ π 0 r 1−α e iθ(1−α) 1 +r 2 e 2iθ · re iθ · idθ ¸ ¸ ¸ ¸ ≤ r 2−α 1 −r 2 π → 0 as r → 0, and _ γ 2 f(z)dz = _ −r −R x 1−α 1 +x 2 dx = _ R r e (1−α) log(−x) 1 +x 2 dx = _ R r x 1−α e i(1−α)π 1 +x 2 dx. Therefore by letting r → 0 and R → ∞, we have _ ∞ 0 x 1−α (1 +e i(1−α)π ) 1 +x 2 dx = _ ∞ 0 x 1−α 1 +x 2 dx · 2 cos (1 −α)π 2 e i(1−α) π 2 = πe i(1−α) π 2 , i.e. _ ∞ 0 x 1−α 1+x 2 dx = π 2 csc _ απ 2 _ . (vi) Proof. This problem is a special of problem (x). See the solution there. (vii) Proof. We use the method outlined in the remark of Problem (ii). _ π 0 dθ a + cos θ = 1 2 _ π −π dθ a + cos θ = 1 2 _ |z|=1 dz iz a + z+z −1 2 = 1 2i _ |z|=1 dz 1 2 z 2 +az + 1 2 . The equation 1 2 z 2 + az + 1 2 = 0 has two roots: z 1 = −a + √ a 2 −1 and z 2 = −a − √ a 2 −1. Clearly, |z 1 | = 1 a+ √ a 2 −1 < 1 and |z 2 | > 1. So _ π 0 dθ a + cos θ = 1 i _ |z|=1 dz (z −z 1 )(z −z 2 ) = 2π · 1 z 1 −z 2 = π √ a 2 −1 . Remark 16. The expansion of integration interval from [0, π] to [−π, π] is motivated by Rule 2. (viii) 44 Proof. Using the result on Dirichlet integral: _ ∞ 0 sin x x dx = π 2 , we have _ ∞ 0 _ sinx x _ 2 dx = _ ∞ 0 sin 2 xd _ − 1 x _ = _ ∞ 0 2 sin xcos xdx x = _ ∞ 0 sin(2x) 2x d(2x) = π 2 . (ix) Proof. It is easy to see that when λ = p = 0, the integral is equal to 1. So without loss of generality, we only consider the cases where λ and p are not simultaneously equal to 0. Choose r, R so that 0 < r < R, and r is sufficiently small and R is sufficiently large. Define γ 1 = {z : r ≤ |z| ≤ R, arg z = 0}, γ 2 = {z : r ≤ |z| ≤ R, arg z = 2π}, γ r = {z : |z| = r, 0 < arg z < 2π}, and γ R = {z : |z| = R, 0 < arg z < 2π}. Let ρ = cos λ, then ρ ∈ (−1, 1]. Finally, define f(z) = z p z 2 +2z cos λ+1 = e p log z z 2 +2ρz+1 . For R sufficiently large and r sufficiently small, the two roots of z 2 + 2rz + 1 = 0 are contained in {z : r < |z| < R}. These two roots are, respectively, z 1 = −ρ + i √ 1 −r 2 = −cos λ + i| sinλ| and z 2 = −ρ−i √ 1 −r 2 = −cos λ−i| sinλ|. So |z 1 | = |z 2 | = 1, and arg z 1 +arg z 2 = 2π. Denote arg z 1 by θ. If λ = 0, we have by Residue Theorem _ γ 1 +γ R −γ 2 −γ r f(z)dz = 2πi[Res(f, z 1 ) + Res(f, z 2 )] = 2πi _ e p log z 1 z 1 −z 2 + e p log z 2 z 2 −z 1 _ = π | sinλ| [e pθi −e p(2π−θ)i ] = − 2πi sinλ sin(λp)e pπi . If λ = 0, we have by Residue Theorem _ γ 1 +γ R −γ 2 −γ r f(z)dz = 2πiRes(f, −1) = −2pπi · e pπi . Meanwhile, we have the estimates ¸ ¸ ¸ ¸ _ γ R f(z)dz ¸ ¸ ¸ ¸ ≤ R p R 2 −2R −1 · R · 2π → 0 as R → ∞, ¸ ¸ ¸ ¸ _ γ r f(z)dz ¸ ¸ ¸ ¸ ≤ r p 1 −2ρr −r 2 · r · 2π → 0 as r → 0, and _ γ 2 f(z)dz = − _ R 0 (xe 2πi ) p dx x 2 + 2ρx + 1 = − _ R 0 x p dx x 2 + 2ρx + 1 e 2pπi . Since 1 −e 2pπi = 2 sin 2 (pπ) −2 sin(pπ) cos(pπ)i = −2i sin(pπ)e pπi , by letting R → ∞ and r → 0, we have _ ∞ 0 x p dx x 2 + 2xcos λ + 1 = _ π sin(λp) sin λsin(pπ) if λ = 0 pπ csc (pπ) if λ = 0 and p = 0. If we take the convention that α sin α = 1 for α = 0, then the above three formulas can be unified into a single one: π sin(λp) sin λsin(pπ) . Remark 17. We choose the above integration path in order to take advantage of the multi-valued function x p (Rule 1). The no symmetry in x 2 + 2xcos λ + 1 is handled by using the full circle instead of half circle. (x) 45 Proof. Choose two positive numbers r and R such that 0 < r < 1 < R. Let γ 1 = {z : r ≤ |z| ≤ R, arg z = 0}, γ 2 = {z : r ≤ |z| ≤ R, arg z = 2π}, γ R = {z : |z| = R, 0 < arg z < 2π} and γ r = {z : |z| = r, 0 < arg z < 2π}. Define f(z) = z 1/p p(z+1)z where z 1/p = e log z p is defined on C \ [0, ∞). Note by substituting y 1 p for x, we get _ ∞ 0 dx 1 +x p = _ ∞ 0 y 1 p dy p(y + 1)y . By Residue Theorem, _ γ 1 +γ R −γ 2 −γ r f(z)dz = 2πRes(f, −1) = 2πi · (−1) 1 p −p = − 2πi p e log e πi p = −2iαe αi , where α = π p . We have the estimates ¸ ¸ ¸ ¸ _ γ R f(z)dz ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ ¸ _ 2π 0 e 1 p log(Re iθ ) p(Re iθ + 1)Re iθ Re iθ · idθ ¸ ¸ ¸ ¸ ¸ ≤ 2πR 1 p p(R −1) → 0 as R → ∞, ¸ ¸ ¸ ¸ _ γ r f(z)dz ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ ¸ _ 2π 0 e 1 p log(re iθ ) p(re iθ + 1)re iθ re iθ · idθ ¸ ¸ ¸ ¸ ¸ ≤ 2πr 1 p p(1 −r) → 0 as r → 0, and _ γ 2 f(z)dz = − _ R r (xe 2πi ) 1 p dx p(x + 1)x = − _ R r x 1 p dx p(x + 1)x · e 2αi . Therefore by letting r → 0 and R → ∞, we have _ ∞ 0 dx 1 +x p = _ ∞ 0 x 1 p dx p(x + 1)x = −2iαe αi 1 −e 2αi = −2iαe αi −2 sin αe ( π 2 +α)i = α sinα = π p csc _ π p _ . Remark 18. The roots of 1 + x p = 0 are not very expressible. So following Rule 1, we make the change of variable x p = y. Then we choose the above integration path to take advantage of the multi-valued function y 1 p . Since there is no symmetry in the denominator (y + 1)y, we used a full circle instead of a half circle. The change of variable x p = y is equivalent to integration on an arc of angle 2π/p instead of 2π. (xi) Proof. (Oops! looks like my solution has a bug. Catch it if you can.) By the change of variable x = 1 y+1 , we have _ 1 0 x 1−p (1 −x) p 1 +x 2 dx = _ ∞ 0 y p dy (y + 1) 3 + (y + 1) . The equation (y + 1) 3 + (y + 1) = 0 has three roots: z 0 = −1, z 1 = −1 + i, and z 2 = −1 − i. Define f(z) = z p (z+1) 3 +(z+1) . Then 2 i=0 Res(f, z i ) = 1 (z + 1)(z −z 1 ) ¸ ¸ ¸ ¸ z=z 2 + 1 (z + 1)(z −z 2 ) ¸ ¸ ¸ ¸ z=z 1 + 1 (z −z 1 )(z −z 2 ) ¸ ¸ ¸ ¸ z=z 0 = 1 −i · (−2i) + 1 i · 2i + 1 −i · i = 0. 46 Let R and r be two positive numbers with R > r. Define γ 1 = {z : r < |z| ≤ R, arg z = 0}, γ 2 = {z : r < |z| ≤ R, arg z = 2π}, γ R = {z : |z| = R, 0 < arg z < 2π}, and γ r = {z : |z| = r, 0 < arg z < 2π}. Then by Residue Theorem _ γ 1 +γ R −γ 2 −γ r f(z)dz = 2πi 2 i=0 Res(f, z i ) = 0. We note ¸ ¸ ¸ ¸ _ Γ R f(z)dz ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ _ 2π 0 (Re iθ ) p (Re iθ + 1) 3 + (Re iθ + 1) Re iθ · idθ ¸ ¸ ¸ ¸ ≤ 2πR p+1 R 3 −3R 2 −4R −2 → 0 as R → ∞, and ¸ ¸ ¸ ¸ _ 2π 0 (re iθ ) p (re iθ + 1) 3 + (re iθ + 1) re iθ · idθ ¸ ¸ ¸ ¸ ≤ 2πr p+1 2 −4r −3r 2 −r 3 → 0 as r → 0. So by letting r → 0 and R → ∞, we have _ ∞ 0 y p (y + 1) 3 + (y + 1) dy + 0 − _ ∞ 0 y p (y + 1) 3 + (y + 1) dy · e 2pπi −0 = 0. Therefore..., “Houston, we got a problem here.” (xii) Proof. Choose r, R such that 0 < r < R and r is sufficiently small and R is sufficiently large. Let γ 1 = {z : r ≤ |z| ≤ R, arg z = 0}, γ 2 = {z : r ≤ |z| ≤ R, arg z = π}, γ R = {z : |z| = R, 0 < arg z < π} and γ r = {z : |z| = r, 0 < arg z < π}. Define f(z) = log z z 2 +2z+2 . By Residue Theorem, _ γ 1 +γ R +γ 2 −γ r f(z)dz = 2πi · Res(f, −1 +i) = π _ log 2 2 + 3 4 πi _ . We note ¸ ¸ ¸ ¸ _ γ R f(z)dz ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ _ 2π 0 log(Re iθ ) R 2 e 2iθ + 2Re iθ + 2 Re iθ · idθ ¸ ¸ ¸ ¸ ≤ log R + 2π R 2 −2R −2 2πR → 0 as R → ∞, ¸ ¸ ¸ ¸ _ γ r f(z)dz ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ _ 2π 0 log(re iθ ) r 2 e 2iθ + 2re iθ + 2 re iθ · idθ ¸ ¸ ¸ ¸ ≤ log r + 2π 2 −2r −r 2 2πr → 0 as r → 0, and _ γ 2 f(z)dz = _ R r log(−x) x 2 −2x + 2 dx = _ R r log x +πi x 2 −2x + 2 dx. Therefore by letting r → 0 and R → ∞, we have _ ∞ 0 _ log x x 2 −2x + 2 + log x x 2 + 2x + 2 _ dx +i _ ∞ 0 πdx x 2 −2x + 2 = π 2 log 2 + 3 4 π 2 i, i.e. _ ∞ 0 _ log x x 2 −2x+2 + log x x 2 +2x+2 _ dx = π 2 log 2. Meanwhile, we note _ ∞ 0 _ log x x 2 −2x+2 − log x x 2 +2x+2 _ dx = _ ∞ 0 log x 4x (x 2 +2) 2 −4x 2 dx = _ ∞ 0 log y y 2 +4 dy. To calculate the value of _ ∞ 0 log y y 2 +4 dy, we apply again Residue Theorem. Let’s define g(z) = log z z 2 +4 and then we have _ γ 1 +γ R +γ 2 −γ r g(z)dz = 2πi · Res(g, 2i) = π 2 _ log 2 + π 2 i _ . We note ¸ ¸ ¸ ¸ _ γ R g(z)dz ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ _ 2π 0 log(Re iθ ) R 2 e 2iθ + 4 Re iθ · idθ ¸ ¸ ¸ ¸ ≤ log R + 2π R 2 −4 2πR → 0 47 as R → ∞, ¸ ¸ ¸ ¸ _ γ r g(z)dz ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ _ 2π 0 log(re iθ ) r 2 e 2iθ + 4 re iθ · idθ ¸ ¸ ¸ ¸ ≤ log r + 2π 4 −r 2 2πr → 0 as r → 0, and _ γ 2 g(z)dz = _ R r log(−x) x 2 + 4 dx = _ R r log x +πi x 2 + 4 dx. So by letting r → 0 and R → ∞, we have 2 _ ∞ 0 log x x 2 + 4 dx +πi _ ∞ 0 dx x 2 + 4 = π 2 log 2 + π 2 4 i. Hence _ ∞ 0 log x x 2 +4 dx = π 4 log 2. Solving the equation _ _ ∞ 0 log x x 2 −2x+2 dx + _ ∞ 0 log x x 2 +2x+2 dx = π 2 log 2 _ ∞ 0 log x x 2 −2x+2 dx − _ ∞ 0 log x x 2 +2x+2 dx = π 4 log 2, we get _ ∞ 0 log x x 2 + 2x + 2 dx = π 8 log 2. Remark 19. Note how we handled the combined difficulty caused by no symmetry in the denominator of the integrand function and log function: because x 2 + 2x + 2 is not symmetric, we want to use full circle; but this will cause the integrals produced by different integration paths to cancel with each other. Therefore we are forced to choose half circle and use two equations to solve for the desired integral. (xiii) Proof. We choose r and R such that 0 < r < R, and r is sufficiently small and R is sufficiently large. Let γ 1 = {z : r ≤ |z| ≤ R, arg z = 0}, γ 2 = {z : r ≤ |z| ≤ R, arg z = π}, γ R = {z : |z| = R, 0 < arg z < π}, and γ r = {z : |z| = r, 0 < arg z < π}. Define f(z) = √ z log z z 2 +1 . Then _ γ 1 +γ R +γ 2 −γ r f(z)dz = 2πiRes(f, i) = π 2 2 ie π 4 i = π 2 2 √ 2 (−1 +i). We note ¸ ¸ ¸ ¸ _ γ R f(z)dz ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ ¸ _ 2π 0 (Re iθ ) 1 2 log(Re iθ ) R 2 e 2iθ + 1 Re iθ · idθ ¸ ¸ ¸ ¸ ¸ ≤ 2πR √ R(log R + 2π) R 2 −1 → 0 as R → ∞, ¸ ¸ ¸ ¸ _ γ r f(z)dz ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ ¸ _ 2π 0 (re iθ ) 1 2 log(re iθ ) r 2 e 2iθ + 1 re iθ · idθ ¸ ¸ ¸ ¸ ¸ ≤ 2πr √ r(log r + 2π) 1 −r 2 → 0 as r → 0, and _ γ 2 f(z)dz = _ R r √ −xlog(−x) x 2 + 1 dx = _ R r √ xe π 2 i (log x +πi) x 2 + 1 dx = e π 2 i _ R r √ xlog x x 2 + 1 dx +e π 2 i πi _ R r √ x x 2 + 1 dx. By letting r → 0 and R → ∞, we have (1 +i) _ ∞ 0 √ xlog x x 2 + 1 dx −π _ ∞ 0 √ x x 2 + 1 dx = π 2 2 √ 2 (−1 +i). Therefore, _ ∞ 0 √ x log x x 2 +1 dx = π 2 2 √ 2 . 48 (xiv) Proof. We note the power series expansion ln(1 −z) = − ∞ n=1 z n n and ln(1 +z) = ∞ n=1 (−1) n+1 z n n hold in |z| < 1 and the convergence is uniform on |z| ≤ 1 −ε, for any ε ∈ (0, 1). Therefore, for any δ > 0, we have _ ∞ δ log _ e x + 1 e x −1 _ dx = _ ∞ δ _ ∞ n=1 (−1) n+1 e −nx n + ∞ n=1 e −nx n _ dx = 2 _ ∞ δ _ ∞ k=1 e −(2k−1)x 2k −1 _ dx = 2 ∞ k=1 e −(2k−1)δ (2k −1) 2 . Therefore _ ∞ 0 log _ e x + 1 e x −1 _ dx = lim δ→0 2 ∞ k=1 e −(2k−1)δ (2k −1) 2 = 2 ∞ k=1 1 (2k −1) 2 , where the last equality can be seen as an example of Lebesgue’s Dominated Convergence Theorem applied to counting measure. By Problem 11 (ii) with z = 1 2 , we know ∞ k=1 1 (2k−1) 2 = π 2 8 . So the integral is equal to π 2 4 . (xv) Proof. Since we have argued rather rigorously in (xiv), we will argue non-rigorously in this problem. _ ∞ 0 x e x + 1 dx = _ ∞ 0 e −x x 1 +e −x dx = _ ∞ 0 e −x x ∞ n=0 (−1) n e −nx dx = ∞ n=0 (−1) n _ ∞ 0 e −(n+1)x xdx = ∞ n=0 (−1) n (n + 1) 2 . Using the function π π 2 sin πz and imitating what we did in Problem 12, we can easily get ∞ n=1 (−1) n−1 n 2 = π 2 12 . Therefore _ ∞ 0 x e x +1 dx = π 2 12 . (xvi) Proof. We provide two solutions, which are essentially the same one. Solution 1. In problem (xvii), we shall prove (let a = 1) _ π 0 xsinx 2 −2 cos x dx = π log 2. Note _ π 0 xsinx 2 −2 cos x dx = _ π 0 x · 2 sin x 2 cos x 2 dx 2 · 2 sin 2 x 2 = 2 _ π 2 0 θ cos θdθ sinθ = −2 _ π 2 0 log(sinθ)dθ. So _ π 2 0 log(sin θ)dθ = − π 2 log. Solution 2. We provide a heuristic proof which discloses the essence of the calculation. Note how Rule 1-3 lead us to this solution. _ π 2 0 log(sinθ)dθ = _ π 2 0 log(cos α)dα = 1 2 _ π 2 − π 2 log(cos α)dα = 1 4 _ π 2 − π 2 log _ 1 + cos 2α 2 _ dα = 1 8 _ π −π log(1 + cos β)dβ − π 4 log 2. (Note how Rule 2 leads us to extended the integration interval from [0, π 2 ] to [−π, π].) Motivated by the similar difficulty explained in the remark of Problem (xvii), we try Rule 1 and note _ π −π log(1 +e iβ )dβ = _ |z|=1 log(1 +z) iz dz = 2π log 1 = 0. 49 For the above application of Cauchy’s integral formula to be rigorous, we need to take a branch of logarithm function that is defined on C \ (−∞, 0] and take a small bypass around 0 for the integration contour, and then take limit. Using the above result, we have 0 = Re _ π −π log(1 +e iβ )dβ = 1 2 _ π −π log _ (1 + cos β) 2 + sin 2 β ¸ dβ = 1 2 _ π −π [log 2 + log(1 + cos β)] dβ. Therefore _ π −π log(1 + cos β)dβ = −2π log 2 and _ π 2 0 log(sinθ)dθ = 1 8 _ π −π log(1 + cos β)dβ − π 4 log 2 = − π 2 log 2. (xvii) Proof. We first assume a > 1. Then by integration-by-parts formula we have _ π 0 xsinx 1 −2a cos x +a 2 dx = 1 2 _ π −π xsinxdx 1 −2a cos x +a 2 = 1 2 _ π −π xd(1 −2a cos x +a 2 ) (1 −2a cos x +a 2 ) · 2a = 1 4a _ 2π log(1 +a) 2 − _ π −π log(1 −2a cos x +a 2 )dx _ . Since a > 1, a − ζ ∈ {z : Rez > 0} for any ζ ∈ ∂D(0, 1). Therefore we can take a branch of the logarithm function such that log(a−z) is a holomorphic function on D(0, 1) and is continuous on ¯ D(0, 1). For example, we can take the branch log z such that it is defined on C \ (−∞, 0] with log(e iπ ) = π and log(e −iπ ) = −π. By Cauchy’s integral formula, _ π −π log(a −e iθ )dθ = _ |z|=1 log(a −z) iz dz = 2π log a. Therefore, _ π −π log(1 − 2a cos x + a 2 )dx = 2Re _ π −π log(a − e iθ )dθ = 4π log a. Plug this equality into the calculation of the original integral, we get π a log _ 1+a a _ . If 0 < a < 1, we have _ π 0 xsinx 1 −2a cos x +a 2 dx = 1 a 2 _ π 0 xsinx 1 − 2 a cos x + 1 a 2 dx = 1 a 2 · π 1 a log _ 1 + 1 a 1 a _ = π a log(a + 1). Assuming the integral as a function of a is continuous at 1, we can conclude for a = 1, the integral is equal to π log 2. The result agrees with that of Mathematica. To prove the continuity rigorously, we split the integral into _ π 2 0 x sin xdx 1−2a cos x+a 2 and _ π π 2 xsin xdx 1−2a cos x+a 2 . For the second integral, we have (x ∈ [ π 2 , π]) xsinx 1 −2a cos x +a 2 = xsinx |a −e ix | ≤ xsinx |a −e i π 2 | ≤ xsinx. So by Lebesgue’s dominated convergence theorem, the second integral is a continuous function of a for a ∈ (0, ∞). For the first integral, we note (1 −2a cos x +a 2 ) takes its minimum at cos x. So xsinx 1 −2a cos x +a 2 ≤ x sinx . Again by Lebesgue’s dominated convergence theorem, the first integral is a continuous function of a for a ∈ (0, ∞). Combined, we conclude the integral _ π 0 xsinx 1 −2a cos x +a 2 dx as a function of a ∈ (0, ∞) is a continuous function. 50 Remark 20. We did not take the transform cos x = z+z −1 2 because log(1 −2a cos x +a 2 ) would then become log[w(a) − w(z)] − log(2a), where w(z) = z+z −1 2 maps D(0, 1) to U := C \ [−1, 1] and we cannot find a holomorphic branch of logarithm function on U. The application of integration-by-parts formula and the extension of integration interval from [0, π] to [−π, π] are motivated by Rule 2. But here a new trick emerged: if the integrand function or part of it could come from the real or imaginary part of a holomorphic function, apply Cauchy’s integral formula (or integration theorem) to that holomorphic function and try to relate the result to our original integral (Rule 1). (xviii) Proof. The function log(z +i) is well-defined on C + := {z : Imz ≥ 0}. We take the branch of log(z +i) that evaluates to log 2 + π 2 i at i. Define γ R = {z : |z| = R, 0 ≤ arg z ≤ π}. Then for R > 1, Residue Theorem implies _ R −R log(z +i) 1 +z 2 dz + _ γ R log(z +i) 1 +z 2 dz = 2πi · i +i i +i = π log 2 + π 2 2 i. Note ¸ ¸ ¸ ¸ _ γ R log(z +i) 1 +z 2 dz ¸ ¸ ¸ ¸ = ¸ ¸ ¸ ¸ _ π 0 log(Re iθ + 1) 1 +R 2 e 2iθ Re iθ · idθ ¸ ¸ ¸ ¸ ≤ πR[log(R + 1) + 2π] R 2 −1 → 0 as R → ∞. So _ ∞ −∞ log(x+i) 1+x 2 dx = π log 2 + π 2 2 i. Hence π log 2 = Re _ ∞ −∞ log(x +i) 1 +x 2 dx = _ ∞ −∞ log √ x 2 + 1 1 +x 2 dx = _ ∞ 0 log(1 +x 2 ) 1 +x 2 dx. Remark 21. The main difficulty of this problem is that the poles of 1 1+x 2 coincide with the branch points of log(x 2 + 1), so that we cannot apply Residue Theorem directly. An attempt to avoid this difficulty might be writing log(x 2 +1) 1+x 2 as the sum of log(x+i) 1+x 2 and log(x−i) 1+x 2 and integrate them separately along different paths which do not contain their respective branch points. But log(x + i) and log(x −i) cannot be defined simultaneously in a region of a Riemann surface which contains both integration paths. But the observation that Re[log(x ±i)] = log √ x 2 + 1 gives us a simple solution (Rule 1). 16. Proof. The power series ∞ n=0 z n is divergent on every point of ∂D(0, 1): if z = 1, ∞ n=0 z n = 1+1+1+· · · ; if z = 1 and z = e iθ , N n=1 z n = N n=0 cos(nθ) +i N n=0 sin(nθ) = sin (N+1)θ 2 cos Nθ 2 sin θ 2 +i sin (N+1)θ 2 sin Nθ 2 sin θ 2 . Similarly, the series ∞ n=0 z −n−2 is divergent on every point of ∂D(0, 1). So the functions represented by these two power series cannot be analytic continuation of each other. 17. Proof. It’s clear we need the assumption that α = 0. The series ∞ n=0 (αz) n is convergent in U 1 = {z : |z| < 1 |α| }. The series 1 1−z ∞ n=0 (−1) n [(1−α)z] n (1−z) n is convergent in U 2 = {z : |1 −α||z| < |1 −z|}. On U 3 = U 1 ∩ U 2 , both of the series represent the analytic function 1 1−αz . So the functions represented by these two series are analytic continuation of each other. 18. 51 Proof. On the line segment {z : 0 < Rez < 1, Imz = 0}, Taylor series of ln(1 + z) is z − 1 2 z 2 + 1 3 z 3 − · · · . On the same line segment, 0 < 1−z 2 < 1 2 . So Taylor series of ln(1 −x) (|x| ≤ 1, x = 1) gives ln2 − 1 −z 2 − (1 −z) 2 2 · 2 2 − (1 −z) 3 3 · 2 3 −· · · = ln2 + ln _ 1 − 1 −z 2 _ = ln(1 +z). Since the two holomorphic functions represented by these two series agree on a line segment, they must agree on the intersection of their respective domains (Theorem 2.13). Therefore they are analytic continuations of each other. 19. Proof. Clearly the series is convergent in D(0, 1) and is divergent for z = 1. So its radius of convergent R = 1. Let f(z) = ∞ n=0 z 2 n . Then f(z) is analytic in D(0, 1) and z = 1 is a singularity of f(z). We note f(z) = z + ∞ n=1 z 2 n = z + ∞ n=1 z 2·2 n−1 = z + ∞ n=1 (z 2 ) 2 n−1 = z +f(z 2 ). Therefore, we have f(z) = z +f(z 2 ) = z +z 2 +f(z 4 ) = z +z 2 +z 4 +f(z 8 ) = · · · . So the roots of equations z 2 = 1, z 4 = 1, z 8 = 1,· · · , z 2 n = 1, · · · , etc. are all singularities of f. These roots form a dense subset of ∂D(0, 1), so f(z) can not be analytically continued to the outside of its circle of convergence D(0, 1). 4 Riemann Mapping Theorem 3. Proof. Let g(z) = ∞ n=0 ¯ c n z n . Then g(z) is holomorphic and has the same radius of convergence as f(z). ∀z ∗ ∈ R ∩ convergence domain of g(z), we have g(z ∗ ) = ∞ n=0 ¯ c n z n ∗ = ∞ n=0 ¯ c n ( ¯ ¯ z ∗ ) n = ¯ f(¯ z ∗ ) = ¯ f(z ∗ ) = f(z ∗ ), where the next to last equality is due to the fact that z ∗ ∈ R and the last equality is due to the fact that f(R) ⊂ R. By Theorem 2.13, f ≡ g in the intersection of their respective domains of convergence. So c n = ¯ c n , i.e. c n (n = 0, 1, · · · ) are real numbers. 11. Proof. Let g(z) = ¯ f(¯ z) (first take conjugate of z, then take conjugate of f(¯ z)). Then ∂g(z) ∂¯ z = ∂ ¯ f(¯ z) ∂¯ z = conjugate _ ∂f(¯ z) ∂z _ = 0. So g is a holomorphic function. Clearly g is univalent. Since U is symmetric w.r.t. real axis, g(U) = ¯ f( ¯ U) = ¯ f(U) = conjugate(D(0, 1)) = D(0, 1). Furthermore, g(z 0 ) = ¯ f(¯ z 0 ) = ¯ f(z 0 ) = 0, and g (z 0 ) = lim z→z 0 ¯ f(¯ z) − ¯ f(¯ z 0 ) z −z 0 = lim z→z 0 conjugate _ f(¯ z) −f(¯ z 0 ) ¯ z − ¯ z 0 _ = conjugate(f (¯ z 0 )) = conjugate(f (z 0 )) = f (z 0 ) > 0. By the uniqueness of Riemann Mapping Theorem, g ≡ f. 52 5 Differential Geometry and Picard’s Theorem 6 A First Taste of Function Theory of Several Complex Variables There are no exercise problems for this chapter. 7 Elliptic Functions There are no exercise problems for this chapter. 8 The Riemann ζ-Function and The Prime Number Theory There are no exercise problems for this chapter. References [1] Johann Boos. Classical and modern methods in summability, Oxford University Press, 2001. [2] John B. Conway. Functions of one complex variable, 2nd Edition, Springer, 1978. [3] Fang Qi-Qin. A course on complex analysis (in Chinese), Peking University Press, 1996. [4] James Munkres. Analysis on manifolds, Westview Press, 1997. [5] M. A. Qazi. The mean value theorem and analytic functions of a complex variable. J. Math. Anal. Appl. 324(2006) 30-38. 53