Chapter 4 Power Series and Laurent Series

Section 4.1 Sequences and Series of Complex Numbers 73

Solutions to Exercises 4.1 1. We have in π e 2 1 0 ≤|an| = = → 0asn →∞. n n

So an → 0 by the squeeze theorem or by applying (2), Sec. 4.1. 5. We have cosh in = cos n (see (25), Sec. 1.6). Hence

cosh in cos n 1 |an| = = ≤ → 0asn →∞. n2 n2 n2

So an → 0 by the squeeze theorem.

9. (a) Suppose that {an} is a convergent sequence and let bn = an+1. We want to show that {bn} is a convergent sequence and that limn→∞ an = limn→∞ bn. Let L denote the limit of the sequence {an}. By deﬁnition of convergence, given >0, we can ﬁnd N ≥ 0, such that for all n ≥ N,

|an − L| <.

But n ≥ N implies that n +1≥ N. So for all n ≥ N,

|an+1 − L| <.

But an+1 = bn, so for all n ≥ N, |bn − L| <, which implies that {bn} is a convergent sequence with same limit as {an}. (b) Deﬁne a = i and a = 3 . Given that {a } is convergent, then by part (a), we have 1 n+1 2+an n limn→∞ an = limn→∞ an+1. Let L = limn→∞ an. Then

3 3 L = lim an = lim an+1 = lim = . n→∞ n→∞ n→∞ 2+an 2+L Solving for L,weﬁnd

3 L = ⇒ L(2 + L)=3; 2+L ⇒ L2 +2L − 3=0, (L + 3)(L − 1)=0,L= −3orL =1.

Because the limit of a convergent sequence is unique, we have to decide whether L =1orL = −3. Note that |a1| = 1. If we can show that Re an ≥ 0, then Re L = limn→∞ Re an ≥ 0, and this would eliminate the value L = −3. Let’s prove by induction that Re an ≥ 0. The statement is clearly true if n = 1, because a1 = i and so Re i =0≥ 0. Now suppose that Re an ≥ 0 and let’s prove that Re an+1 ≥ 0. For this purpose, we compute

3 2+an ! Re an+1 =Re =3Re 2+an (2 + an) · (2 + an)

3 2 = · Re (2 + Re an − i Im an)=3α (2 + 2 Re an) ≥ 0, (2 + an) · (2 + an)

2 1 1 where α = = 2 . This completes the proof by induction and shows that L =6 −3, (2+an )·2+an |2+an | so L =1. 74 Chapter 4 Power Series and Laurent Series

∞ 3 − i 13. The series X is a constant multiple of a convergent geometric series and so it is (1 + i)n n=3 convergent. To ﬁnd its sum proceed as follows:

∞ 3 − i 1 1 X =(3− i) + + ··· (1 + i)n (1 + i)3 (1 + i)4 n=3 1 1 1 =(3− i) 1+ + + ··· (1 + i)3 1+i (1 + i)2 ∞ 3 − i 1 3 − i 1 = X = · , (1 + i)3 (1 + i)n (1 + i)3 1 − r n=0

where r = 1 , |r| = √1 < 1. So the sum is 1+i 2 3 − i 1 3 − i −1 − 3i 3 i S = · = = = − + . (1 + i)3 − 1 (1 + i)2 i (1 + i)2 2 2 1 1+i

∞ 1 1 17. The series X is absolutely convergent by comparison to the series 2 P 2 : (n + i)((n − 1) + i) n n=2 1 |(n + i)((n − 1) + i)| = (n + i) (n − 1) + i ≥ n(n − 1) ≥ n2 if n ≥ 2; 2 so 1 2 ≤ . (n + i)((n − 1) + i) n2

To sum the series, use partial fractions and the terms will telescope, as follows: 1 1 1 = − . (n + i)((n − 1) + i) (n − 1) + i) n + i

So the Nth partial sum is

N 1 1 s = X + N (n − 1) + i) n + i n=2 1 1 1 1 1 1 = − + − + ···+ − (2 − 1) + i) 2+i (3 − 1) + i) 3+i (N − 1) + i) N + i 1 1 1 = − → − 0, 2+i N + i 1+i →∞ 1 1−i as N . So the sum is S = 1+i = 2 . ∞ 1+3i n √ 21. The series X is a geometric series with z = 1+3i . Since |z| = 1 10 < 1, the series 4 4 4 n=0 converges to 1 1 4 4 1 2 = = = = (1 + i). 1 − z − 1+3i 3 − 3i 3 1 − i 3 1 4 25. Write (1+2in)n 1 n = +2i . nn n Section 4.1 Sequences and Series of Complex Numbers 75

1 | |≥| | It is now clear that the series is a geometric series with z = n +2i. Since z Im z = 2, it follows that the series diverges. 29. For n ≥ 2, we have

n −n cos(in) cosh n e + e 1 3 3 3 = = = (e−n +n + e−n −n) ≤ e−n /2. en3 en3 2en3 2 ∞ 3 The series is absolutely convergent by comparison with the convergent series X e−n /2. n=1 33. The series converges as long as

z < 1 ⇒|z| < 2. 2 37. The series converges as long as

1 < 1 ⇒ 1 < |2 − 10z| 2 − 10z

1 1 ⇒ 1 < | − z| 10 5 1 ⇒ 10 < | − z|. 5 1 Thus the series converges outside the closed disk of radius 10 and center at 5 . The limit in this region is (use the geometric series minus the ﬁrst term, which is 1) 1 2 − 10z 1 − 1= − 1= . − 1 − − 1 2−10z 1 10z 1 10z

i 41. If the nth partial sum of the series is sn = n , then the series converges and its limit is i lim sn = lim =0. n→∞ n→∞ n An example of such a series is ∞ i i 1+X − . n +1 n n=1 45. The test of convergence in Theorem 12 is really about series with positive real terms. For a proof, see any calculus book and use the same proof for real series. 49. To establish the addition formula for cosines, we will manipulate the series for cosine, using several properties of absolutely convergent series and Cauchy products. We have

∞ ∞ z2n z2n cos z = X(−1)n 1 ; cos z = X(−1)n 2 ; 1 (2n)! 2 (2n)! n=0 n=0 n z2k z2(n−k) cos z cos z = X(−1)k 1 (−1)n−k 2 1 2 (2k)! (2(n − k))! k=0 (−1)n n (2n)! = X z2kz2(n−k) (2n)! (2k)!(2(n − k))! 1 2 k=0 n n (−1) 2n − = X z2kz2(n k). (2n)! 2k 1 2 k=0 76 Chapter 4 Power Series and Laurent Series

Similarly,

∞ ∞ z2n+1 z2n+1 sin z = X(−1)n 1 ; sin z = X(−1)n 2 ; 1 (2n + 1)! 2 (2n + 1)! n=0 n=0 n z2k+1 z2(n−k)+1) sin z sin z = X(−1)k 1 (−1)n−k 2 1 2 (2k + 1)! (2(n − k) + 1)! k=0 (−1)n n (2(n + 1))! = X z2k+1z2(n+1)−(2k+1) (2(n + 1))! (2k)!(2(n +1)− (2k + 1)))! 1 2 k=0 n n (−1) 2(n +1) − = X z2k+1z2(n+1) (2k+1). (2(n + 1))! 2k +1 1 2 k=0 Hence

∞ n n (−1) 2n − cos z cos z − sin z sin z = X X z2kz2(n k) 1 2 1 2 (2n)! 2k 1 2 n=0 k=0 ∞ n n (−1) 2(n +1) − − X X z2k+1z2(n+1) (2k+1) (2(n + 1))! 2k +1 1 2 n=0 k=0 ∞ n n (−1) 2n − =1+X X z2kz2(n k) (2n)! 2k 1 2 n=1 k=0 ∞ n−1 n−1 (−1) 2n − − X X z2k+1z2n (2k+1), (2n)! 2k +1 1 2 n=1 k=0 where in the ﬁrst series on the right, we wrote the ﬁrst term separately, and in the second series on the right, we shifted the index by changing n to n − 1. Combining the two series, we see that the last displayed sum equals

∞ n n n−1 (−1) ( 2n − 2n − ) 1+X X z2kz2(n k) + X z2k+1z2n (2k+1) . (2n)! 2k 1 2 2k +1 1 2 n=1 k=0 k=0 But the two inner sums in k add up to

2n 2n X zkz2n−k =(z + z )n , k 1 2 1 2 k=0 because the ﬁrst sum adds the even-indexed terms and the second sum adds the odd-indexed terms. Hence ∞ (−1)n cos z cos z − sin z sin z =1+X (z + z )n 1 2 1 2 (2n)! 1 2 n=1 ∞ (−1)n = X (z + z )n = cos(z + z ), (2n)! 1 2 1 2 n=0 as desired. Note: While this was a good exercise in Cauchy products, it is not the most eﬃcient way to prove the addition formula for the cosine. For a more elegant proof, see Example 2(b), Sec. 4.6. Section 4.2 Sequences and Series of Functions 77

Solutions to Exercises 4.2 sin nx 1. The sequence of functions f (x)= converges uniformly on the interval 0 ≤ x ≤ π. To see n n this, let Mn = max |0 − fn(x)| = max |fn(x)|, where the maximum is taken over all x in [0,π]. Then ≤ 1 → →∞ Mn = n . Since Mn 0, as n , it follows that fn converges uniformly to f =0on[0,π]. In fact, we fact uniform convergence on the entire real line.

5. (a) and (b) First, let us determine the pointwise limit of the sequence of functions fn(x)= nx .Forx in the interval 0 ≤ x ≤ 1, we have n2x2 − x +1 nx nx x f (x)= = = → 0, as n →∞. n 2 2 − 2 2 − x 1 2 − x 1 n x x +1 n (x n2 + n2 n(x n2 + n2 Does the sequence converge to 0 uniformly for all x in [0, 1]? To answer this question we estimate the maximum possible diﬀerence between 0 and fn(x), as x varies in [0, 1]. For this purpose, we compute Mn = max |fn(x)| for x in [0, 1]. We have

3 2 0 n − n x 0 3 2 1 fn(x)= ; fn(x)=0⇒ n − n x =0⇒ x = ; (1 − x + n2 x2)2 n 1 1 1 1 f ( )= > ⇒ M > . n n − 1 2 n 2 2 n

Since Mn does not converge to 0, we conclude that the sequence does not converge uniformly to 0 on [0, 1]. (c) The sequence does converge uniformly on any interval of the form [a, b], where 0

(b) Let Mn = max |fn(z)| for |z|≤1, To prove that fn converges uniformly, we must show that Mn → 0, as n →∞. We have 1 1 1 z + |z| + 1+ |f (z)| = n ≤ n ≤ n . n z |z| 1 n( 2 +2) − n(2 − 2 ) n n(2 n2 ) n

This shows that Mn is smaller than the last displayed expression, which tends to 0 as n →∞. Hence fn does converge to 0 uniformly for all |z|≤1. 13. If |z|≤1, then zn 1 ≤ . n(n +1) n(n +1)

1 Apply the Weierstrass M-test with Mn = n(n+1) . Since P Mn is convergent, it follows that the ∞ zn series X , converges uniformly for all |z|≤1. n(n +1) n=1 17. If |z|≤2, then n n z +2 4 ≤ . 5 5

78 Chapter 4 Power Series and Laurent Series

4 n Apply the Weierstrass M-test with Mn = 5 . Since P Mn is convergent (a geometric series with ∞ z +2 n r<1), it follows that the series X converges uniformly for all |z|≤2. 5 n=0 21. If 2.01 ≤|z − 2|≤2.9, then

n (z − 2)n 2.9 ≤ = An, 3n 3

and n 2n 2 ≤ = Bn. (z − 2)n 2.01

Apply the Weierstrass M-test with Mn =(An + Bn). Since P Mn is convergent (two geometric ∞ (z − 2)n 2n series with ratios < 1), it follows that the series X + converges uniformly in 3n (z − 2)n n=0 the annular region 2.01 ≤|z − 2|≤2.9. | − 1 | 1 25. (a) If z 2 < 6 then 1 1 1 1 1 1 2 |z|≤|z − + |≤|z − | + < + = < 1. 2 2 2 2 6 2 3 ∞ n | − 1 | 1 n So the series Pn=0 z converge uniformly on z 2 < 6 by the Weierstrass M-test with Mn = (23) . | − 1 | 1 ∞ n (b) If z 2 < 2 , then z can get very close to the value 1, where the series Pn=0 z does not | − 1 | 1 converge. So the initial guess is that the series is not uniformly convergent in the region z 2 < 2 . To prove this assertion, you can repeat the proof given in Example 4; in particular, the inequalities in (3) and the argument that follows them still hold. 29. (a) Let δ>1 be a positive real number. To show that the series

∞ 1 ζ(z)=X (principal branch of nz) nz n=1

converges uniformly on the half-plane Hδ = {z :Rez ≥ δ>1}, we will apply the Weierstrass M-test. For all z ∈ Hδ, we have

z (x+iy)lnn x ln n x δ |n | = e = e = n >n .

So 1 1 ≤ = Mn. nz nδ

1 Since P Mn = P nδ is a convergent series (because δ>1), it follows from the Weierstrass M-test ∞ 1 that that Pn=1 nz converges uniformly in Hδ. ∞ 1 { } (b) Each term of the series Pn=1 nz is analytic in H = z :Rez>1 (in fact, each term is entire). To conclude that the series is analytic in H, it is enough by Corollary 2 to show that the series converges uniformly on any closed disk contained in H.IfS is a closed disk contained in H, S is clearly disjoint from the imaginary axis. Let Hδ (δ>0) be a half-plane containing S. By part (a), the series converges uniformly on Hδ, consequently, the series converges uniformly on S.By Corollary 2, the series is analytic in H. (Note the subtilty in the proof. We did not show that the series converges uniformly on H. In fact, the series does not converge uniformly in H.) (c) To compute ζ0(z), according to Corollary 2, we can diﬀerentiate the series term-by-term. Write 1 1 = = e−z ln n. nz ez ln n Section 4.2 Sequences and Series of Functions 79

Using properties of the exponential function, we have d 1 d 1 = e−z ln n = − ln ne−z ln n = − ln n . dz nz dz nz So, for all z ∈ H, ∞ ln n ζ0(z)=− X . nz n=1

33. To show that fn {fn} converges uniformly on Ω, it is enough to show that

max |fn(z) − fm(z)| z∈Ω can be made arbitrarily small by choosing m and n large. In other words, given >0, we must show that there is a positive integer N such that if m, n ≥ N, then

max |fn(z) − fm(z)| <. z∈Ω

This will show that the sequence {fn} is uniformly Cauchy, and hence it is uniformly convergent by Exercise 30. Since each fn is analytic inside C and continuous on C, it follows that fn − fm is also analytic inside C and continuous on C. Since C is a simple closed path, the region interior to C is a bounded region. By the maximum principle, Corollary 2, Sec. 3.7, the maximum value of |fn − fm| occurs on C. But on C the sequence {fn} is a Cauchy sequence, so there is a positive integer N such that if m, n ≥ N, then max |fn(z) − fm(z)| <. z∈C Hence max |fn(z) − fm(z)|≤max |fn(z) − fm(z)| <, z∈Ω z∈C which is what we want to prove. The key idea in this exercise is that the maximum value of an analytic function occurs on the boundary. So the uniform convergence of a sequence inside a bounded region can be deduced from the uniform convergence of the sequence on the boundary of the region. 80 Chapter 4 Power Series and Laurent Series

Solutions to Exercises 4.3 1. Apply the ratio test: For z =0,6

n+1 an+1 z 2n +1 ρ = lim = lim (−1)n+1 · (−1)n n→∞ a n→∞ 2n +1+1 zn n 2n +1 n(2 + 1 ) = |z| lim = |z| lim n = |z|. n→∞ n→∞ 3 2n +3 n(2 + n )

Thus the series converges absolutely if |z| < 1 and diverges if |z| > 1. The radius of convergence is 1, the disk of convergence is |z| < 1, the circle of convergence is |z| =1. ∞ (4iz − 2)n 5. Apply the ratio test: For z =6 2 , X . 4i 2n n=0

n+1 n an+1 (4iz − 2) 2 ρ = lim = lim · n→∞ a n→∞ 2n+1 (4iz − 2)n n |4iz − 2| = . 2

Thus the series converges absolutely if

|4iz − 2| 2 1 2 (−i) 1 < 1 ⇔|4i(z − )| < 2 ⇔|z − (−i)| < ⇔|z − | < . 2 4i 2 4 2 2

| − (−i) | 1 1 The series diverges if z 2 > 2 . The radius of convergence is 2 ; the disk of convergence is | − (−i) | 1 | − (−i) | 1 z 2 < 2 ; and the circle of convergence is z 2 = 2 . 9. We compute the radius of convergence by using the Cauchy-Hadamard formula

1 n in π in π = limsup q (1 − e 4 )n = lim sup 1 − e 4 =2. R

To understand why the limsup is equal to 2, recall that the limsup is the limit of the sup of the in π ∞ in π tail of the sequence { 1 − e 4 } ,asN tends to ∞. The terms e 4 take values from the set √ √ n=N 2 2 in π {± ± ± ± } − 4 2 2 , i, 1 . So the largest value of 1 e is 2, and this value repeats inﬁnitely often, 1 which explains the value of the limsup. Thus, R = 2 . 13. Start with the geometric series

∞ 1 = X zn, |z| < 1. 1 − z n=0

Diﬀerentiate term-by-term: ∞ 1 = X nzn−1, |z| < 1. (1 − z)2 n=1 Multiply both sides by 2: ∞ 2 = X 2nzn−1, |z| < 1. (1 − z)2 n=1 Section 4.3 Power Series 81

17. We have ∞ ∞ ∞ (3z − i)n (3(z − i ))n i X = X 3 = X(z − )n 3n 3n 3 n=0 n=0 n=0 ∞ i = X wn (w = z − ) 3 n=0 1 1 i = = z − < 1 1 − w 1 − (z − i ) 3 3 3 = , 3+i − 3z which is valid for z − i < 1. 3 21. This exercise is a simple application of Theorem 15, Sec. 4.1, and basic properties of power ∞ n ∞ n series. If Pn=0 an(z − z0) has radius of convergence R1 > 0 and Pn=0 bn(z − z0) has radius of convergence R2 > 0, then these series converge absolutely within their radii of convergence. Apply Theorem 15, Sec. 4.1: If z is inside the disk of convergence of both series; that is, if |z − z0|

Thus, for |z − z0|

(d) Switching to polar coordinates: u = r cos θ, v = r sin θ, u2 + v2 = r2, dudv = rdrdθ; for (u, v) ≤ ∞ ≤ ∞ ≤ ≤ π ≤ ∞ varying in the ﬁrst quadrant (0 u< and 0 v< ), we have 0 θ 2 ,and0 r< , and the double integral in (c) becomes

∞ π 2 2 Z Z −r 2z1 −1 2z2−1 Γ(z1)Γ(z2)=4 e (r cos θ) (r sin θ) rdrdθ 0 0

=Γ(z1+z2 ) π ∞ 2 z }| 2 { =2Z (cos θ)2z1 −1(sin θ)2z2 −1dθ 2 Z r2(z1+z2 )−1e−r dr 0 0 (use (b) with z1 + z2 in place of z) π 2 Z 2z1−1 2z2−1 = 2Γ(z1 + z2) (cos θ) (sin θ) dθ, 0 implying (d). Section 4.4 Taylor Series 83

Solutions to Exercises 4.4

1. According to Theorem 1, the Taylor series around z0 converges in the largest disk, centered at z−1 z0 = 0, in which the function is analytic. Clearly, e is entire, so radius of convergence is R = ∞.

5. According to Theorem 1, the Taylor series around z0 converges in the largest disk, centered at z +1 z =2+i, in which the function is analytic. The function f(z)= is analytic for all z =6 i.So 0 z − i the largest disk around z0 on which f is analytic has radius R = |z0 − i| = |2+i − i| =2.

9. According to Theorem 1, the Taylor series around z0 converges in the largest disk, centered at 6 π z0 = 0, in which the function is analytic. The function f(z) = tan z is analytic for all z = 2 +2kπ. So the largest disk around z0 on which f is analytic has radius equal to the distance from z0 =0to | − π | π the nearest point where f fails to be analytic. Clearly, then R = 0 2 = 2 . z 13. Arguing as we did in Exercises 1-9, we ﬁnd that the Taylor series of f(z)= around 1 − z z0 = 0 has radius of convergence equal to the distance from z0 = 0 to the nearest point where f fails to be analytic. Thus R = 1. (This will also come out of the computation of the Taylor series.) Now, for |z| < 1, the geometric series tells us that ∞ 1 = X zn. 1 − z n=0 Multiplying both sides by z, we get, for |z| < 1, ∞ z = X zn+1. 1 − z n=0 17. Because the function is entire, the Taylor series will have an inﬁnite radius of convergence. Note that the series expansion around 0 is easy to obtain: ∞ ∞ ∞ zn zn zn+1 ez = X ⇒ zez = z X = X . n! n! n! n=0 n=0 n=0

But how do we get the series expansion around z0 = 1? In the previous expansion, replacing z by z − 1, we get ∞ (z − 1)n+1 (z − 1)ez−1 = X . n! n=0 The expansion on the right is a Taylor series centered at z0 = 1, but the function on the left is not quite the function that we want. Let f(z)=zez. We have (z − 1)ez−1 = e−1zez − ez−1 = e−1f(z) − ez−1. So f(z)=e(z − 1)ez−1 + ez−1. Using the expansion of (z − 1)ez−1 and the expansion of ez−1 = ∞ (z−1)n Pn=0 n! ,weﬁnd ∞ ∞ ∞ ∞ (z − 1)n+1 (z − 1)n (z − 1)n (z − 1)n f(z)=e X + X = e X + X n! n! (n − 1)! n! n=0 n=0 n=1 n=0 ∞ ∞ (z − 1)n (z − 1)n = e X ++1+X (n − 1)! n! n=1 n=1 ∞ ∞ 1 1 n 1 = e1+X(z − 1)n + = e1+X(z − 1)n + (n − 1)! n! n! n! n=1 n=1 ∞ n +1 = e1+X(z − 1)n . n! n=1 84 Chapter 4 Power Series and Laurent Series

21. (a) To obtain the partial fractions decomposition 1 1 1 = − , (1 − z)(2 − z) 1 − z 2 − z we proceed in the usual way: 1 A B = + (1 − z)(2 − z) 1 − z 2 − z A(2 − z)+B(1 − z) = ; (1 − z)(2 − z) 1=A(2 − z)+B(1 − z) Take z =2 ⇒ 1=−B, B = −1. Take z =1 ⇒ 1=A.

Thus we obtain the desired partial fractions decomposition. Expanding each term in the partial fractions decomposition around z0 = 0, we obtain ∞ 1 = X zn, |z| < 1; 1 − z n=0 ∞ 1 1 1 z n z − = − = − X , | | < 1, or |z| < 2. 2 − z 2(1 − z ) 2 2 2 2 n=0 So, for |z| < 1, ∞ 1 1 1 1 = − = X(1 − )zn. (1 − z)(2 − z) 1 − z 2 − z 2n+1 n=0 (b) We an derive the series in (a) by considering the Cauchy products of the series expansions of 1 1 1−z and 2−z , as follows. From (a), we have ∞ ∞ ∞ 1 1 zn · = X zn · X = X c zn, 1 − z 2 − z 2n+1 n n=0 n=0 n=0

where cn is obtained from the Cauchy product formula (see Exercise 21, Sec. 4.3):

n cn = X akbn−k, k=0 1 a =1,b− = , k n k 2n−k+1 n 1 1 n c = X = X 2k. n 2n−k+1 2n+1 k=0 k=0 (c) To show that the Cauchy product is the same as the series that we found in (a), we must prove that 1 n 1 X 2k =(1− ). 2n+1 2n+1 k=0 But this is clear since n X 2k =1+2+22 + ···+2n =2n+1 − 1, k=0 Section 4.4 Taylor Series 85

and so 1 n 1 1 X 2k = (2n+1 − 1) = (1 − ). 2n+1 2n+1 2n+1 k=0 The radius of the Maclaurin series is 1. This follows from our argument in (a) or from Theorem 1, since the function has a problem at z =1. 1 25. The easiest way to do this problem is to start with the series expansion of f(z)= and z − 2i then differentiate it term-by-term, twice. Let’s see: 1 1 −i i 1 = = = · − z − z − − z z 2i 2i( 2i 1) 2( 2i 1) 2 1 2i ∞ ∞ n i z n i z = X = X(−i)n , 2 2i 2 2n n=0 n=0 | z | | | which is valid for 2i < 1or z < 2. Within the radius of convergence, the series can be differentiated term by term as often as we wish. Differentiating once, we obtain

∞ −1 i zn−1 = X(−i)nn . (z − 2i)2 2 2n n=1 Diﬀerentiating a second time, we obtain

∞ ∞ 2 i zn−2 1 i zn−2 = X(−i)nn(n − 1) ⇒ = X(−i)nn(n − 1) . (z − 2i)3 2 2n (z − 2i)3 4 2n n=2 n=2 All series are valid in |z| < 2. If we shift the index of summation of the series (change n to n + 2), we obtain the series ∞ ∞ i zn i zn X(−i)n+2(n + 2)(n +1) = − X(−i)n(n + 2)(n +1) . 4 2n+2 4 2n+2 n=0 n=0

29. For all z, we have

∞ 2 n ∞ 2n 4 2 (z ) z z ez = X = X =1+z2 + + ···. n! n! 2! n=0 n=0 Hence, for all z, 2 2 z ez − 1=z2 + + ··· . 2! If z =6 0, we have 4 6 2 4 2 z z z z ez − 1=z2 + + + ···= z2(1 + + + ···), 2! 3! 2! 3! so, if z =0,6 2 2 4 ez − 1 z2(1 + z + z + ···) z2 z4 = 2! 3! =1+ + + ···, z2 z2 2! 3! where the series converges for all z. But a power series is analytic in the disk where it converges. So z2 z4 ··· ∞ z2n ∞ z2n the series 1 + 2! + 3! + = Pn=0 (n+1)! is entire. Call g(z)=Pn=0 (n+1)! . We just proved that 2 6 ez −1 02 04 ··· for z =0,g(z)= z2 = f(z). Now for z = 0, we have g(0) = 1 + 2! + 3! + =1=f(0). Thus f(z)=g(z) for all z and since g is entire, it follows that f is entire and its Maclaurin series is g(z). 86 Chapter 4 Power Series and Laurent Series

33. (a) The sequence of integers {ln} satisﬁes the recurrence relation ln = ln−1 + ln−2 for n ≥ 2, with l0 =1andl1 = 3. As suggested, suppose that ln occur as the Maclaurin coeﬃcient of some ∞ n analytic function f(z)=Pn=0 lnz , |z|

1 1 1 1 1 " 1 1 # = − = z − z . (z1 − z)(z2 − z) z1 − z2 z2 − z z1 − z z1 − z2 z2(1 − ) z1(1 − ) z2 z1 Apply a geometric series expansion and simplify:

∞ ∞ 1 1 " 1 z n 1 z n# = X − X (z − z)(z − z) z − z z z z z 1 2 1 2 2 n=0 2 1 n=0 1 ∞ 1 1 1 = X − zn z − z n+1 n+1 1 2 n=0 z2 z1 ∞ 1 (zn+1 − zn+1) = X 1 2 zn. z − z (z z )n+1 1 2 n=0 1 2 Section 4.4 Taylor Series 87

Now, consider the function 1 −1 −1 = = , 2 2 1 − z − z z + z − 1 (z1 − z)(z2 − z)

2 where z1 and z2 are the roots of z + z − 1: √ √ −1+ 5 −1 − 5 z = and z = , 1 2 2 2 arranged so that z1| < |z2|. These roots satisfy known relationships determined by the coeﬃcients of the polynomial z2 + z − 1. We will need the following easily veriﬁed identities: √ z1 − z2 = 5 and z1z2 = −1.

We will also need the following identities:

√ n √ n n 1 − 5! −1+ 5 z (z1 − 2) = (−1) − 2+ 1 2 2 √ n √ 1 − 5! −5+ 5 =(−1)n 2 2 √ n+1 √ 1 − 5! =(−1)n 5 . 2

Similarly,

√ n √ n n 1+ 5! 1+ 5 z (2 − z2)=(−1) 2+ 2 2 2 √ n √ 1+ 5! 5+ 5 =(−1)n 2 2 √ n √ √ n+1 1+ 5! √ 1+ 5 √ 1+ 5! =(−1)n 5 =(−1)n 5 . 2 2 2

We are now ready to derive the desired Maclaurin series. We have 1 −1 −1 = = 2 2 1 − z − z z + z − 1 (z1 − z)(z2 − z) ∞ ∞ −1 zn+1 − zn+1 −1 = √ X 1 2 zn = √ X(−1)n+1zn+1 − zn+1zn (−1)n+1 1 2 5 n=0 5 n=0 √ = 5 ∞ −1 −1 n+1 n+1 n+1 n = √ z(z1 }|− z2)({ −1) + √ X(−1) z1 − z2 z 5 5 n=1 ∞ 1 n n+1 n+1 n =1+√ X(−1) z1 − z2 z ; 5 n=1 ∞ ∞ 2z 2 −2 = √ X(−1)nzn+1 − zn+1zn+1 = √ X(−1)nzn − znzn. 1 − z − z2 1 2 1 2 5 n=0 5 n=1 88 Chapter 4 Power Series and Laurent Series

So 1+2z f(z)= 1 − z − z2 ∞ 1 n n n n+1 n+1 n =1+√ X(−1) 2z2 − 2z1 − z2 + z1 z 5 n=1 ∞ 1 n n n n =1+√ X(−1) z2 (2 − z2)+z1 (z1 − 2)z 5 n=1 ∞ √ n+1 √ n+1 1  √  1+ 5! 1 − 5!   =1+√ X (−1)n(−1)n 5 + zn 5  2 2  n=1   ∞ √ n+1 √ n+1  1+ 5! 1 − 5!  =1+X + zn.  2 2  n=1

Thus √ n+1 √ n+1 1+ 5! 1 − 5! l = + ,n≥ 0. n 2 2

1 | | 37. We use the binomial series expansion from Exercise 36, with α = 2 . Accordingly, for z < 1,

∞ 1 1 n (1 + z) 2 = X 2 z , n n=0 where, for n ≥ 1,

1 1 ( 1 − 1) ···( 1 − n +1) 1 −3 ···−(2n−3) 2 = 2 2 2 = 2 2 2 n n! n! − 1 3 ···(2n 3) 1 1 · 3 · 5 ···(2n − 3) · 2 · 4 ···(2n − 2) =(−1)n−1 2 2 2 =(−1)n−1 n! 2nn! 2 · 4 ···(2n − 2) 1 (2n − 2)! 1 (2n − 2)! =(−1)n−1 =(−1)n−1 2nn! 2 · 1 · 2 · 2 ···2 · (n − 1) 2nn! 2n−1 · 1 · 2 ···(n − 1) 1 (2n − 2)! 1 (2n − 2)! 2n(2n − 1) ==(−1)n−1 =(−1)n−1 2nn! 2n−1(n − 1)! 2nn! 2n−1(n − 1)! 2n(2n − 1) 1 (2n)! (2n)!(2n − 1) (−1)n−1 2n =(−1)n−1 (−1)n−1 =(−1)n−1 . 2nn! 2nn!(2n − 1) 22n(n!)2 22n(2n − 1) n

Thus ∞ n−1 1 (−1) 2n n (1 + z) 2 = X z , |z| < 1. 22n(2n − 1) n n=0

41. (a) and (b) There are several possible ways to derive the Taylor series√ expansion of f(z) = Log z about the point z0 = −1+i. Here is one way. Let z0 = −1+i,so|z0| = 2. The function Log z is analytic except on the negative real axis and 0. So it is guaranteed by Theorem 1 to have a series expansion in the largest disk around z0 that does not intersect the negative real axis. Such a disc, as you can easily verify, has radius Im z0 = 1. However, as√ you will see shortly, the series that we obtain has a larger radius of convergence, namely |z0| = 2 (of course, this is not a contradiction to Theorem 1). Section 4.4 Taylor Series 89

Consider the function Log z in B1(z0), where it is analytic. (The disk of radius 1, centered at z0 ∈ d 1 is contained in the upper half-plane.) For z B1(z0), we have dz Log z = z . Instead of computing 1 the Taylor series of Log z directly, we will ﬁrst compute the Taylor series of z , and then integrate term-by-term within the radius of convergence of the series (Theorem 3, Sec. 4.3). Getting ready to apply the geometric series result, we write 1 1 1 = = (z0 =0)6 z0−z z z0 − (z0 − z) z0 1 − z0 ∞ n 1 1 1 z0 − z = · z −z = X z0 1 − 0 z0 z0 z0 n=0 ∞ n 1 z − z0 = X(−1)n , z zn 0 n=0 0 where the series expansion holds for

z0 − z < 1 ⇔|z − z| < |z |. z 0 0 0 √ Thus the series representation holds in a disk of radius |z0| = 2, around z0. Within this disk, we can integrate the series term-by-term and get

z ∞ n z ∞ n Z 1 1 (−1) Z n 1 (−1) n+1 dζ = X ζ − z0 dζ = X z − z0 . ζ z zn z (n +1)zn z0 0 n=0 0 z0 0 n=0 0 Reindexing the series by changing n +1ton, we obtain

z ∞ n+1 √ Z 1 (−1) n dζ = X z − z0 |z − z0| < |z0| = 2. ζ nzn z0 n=1 0

1 Now we have to decide what to write on the left side. The function Log z is an antiderivative of z in the disk of radius 1, centered at z0. (Remember that Log z is not analytic on the negative real axis, so we cannot take a larger disk.) So, for |z − z0| < 1, we have

z Z 1 z dζ = Log ζ = Log z − Log z0. ζ z z0 0

Thus, for |z − z0| < 1, we have

∞ n+1 (−1) n Log z = Log z0 + X z − z0 , nzn n=1 0 √ even though the series on the right converges in the larger disk |z − z0| < 2. 90 Chapter 4 Power Series and Laurent Series

Solutions to Exercises 4.5 1. Apply (5) with w = −z2 and get

∞ 1 1 = − X , 1 < |z2|; 1+z2 (−z2)n n=1 equivalently, ∞ 1 (−1)n+1 = X , 1 < |z|. 1+z2 z2n n=1 1 It is worth doing this problem without appealing to formula (5). Since z2 < 1, we can use a 1 geometric series in z2 as follows. We have

Expand using a geometric series. 1 1 z }|1 { = 1+z2 z2 − −1 1 z2 ∞ ∞ ∞ 1 −1 1 (−1)n (−1)n = X( )n = X = X z2 z2 z2 z2n z2(n+1) n=0 n=0 n=0 ∞ (−1)n−1 = X , z2n n=1

where in the last series we shifted the index of summation by 1. Note that (−1)n−1 =(−1)n+1, and so the two series that we derived are the same. 5. We have ∞ (−1)n−1 Log (1 + w)=X wn, |w| < 1. n n=1 1 Put w = z , then ∞ n−1 1 (−1) 1 n 1 Log (1 + )=X , | | < 1; z n z z n=1 equivalently ∞ 1 (−1)n−1 1 Log 1+ = X , 1 < |z|. z n zn n=1 9. Simplify the function by using sin 2a = 2 sin a cos a, and get

sin 1 cos 1 1 1 2 1 2 z z = sin = sin . z z 2 z 2z z Start with the Maclaurin series for sin w: for all w,

∞ (−1)n sin w = X w2n+1. (2n + 1)! n=0 6 2 For z = 0, put w = z ,

∞ n ∞ n 2n+1 2 (−1) 2 2n+1 (−1) 2 sin = X = X . z (2n + 1)! z (2n + 1)!z2n+1 n=0 n=0 Section 4.5 Laurent Series 91

1 6 To get the desired expansion, multiply by 2z : for z =0,

∞ ∞ 1 2 1 (−1)n22n+1 (−1)n22n sin = X = X . 2z z 2z (2n + 1)!z2n+1 (2n + 1)!z2(n+1) n=0 n=0

13. Applying a partial fractions decomposition, we ﬁnd that z 3 2 = − . (z + 2)(z +3) z +3 z +2

In the annulus 2 < |z| < 3, we have 2 < 1 and z < 1. So to expand 3 , factor the 3 in the z 3 z+3 denominator and you’ll get 3 3 1 = = , z − −z z +3 3(1 + 3 ) 1 ( 3 ) where −z < 1. So we can apply a geometric series expansion and obtain: 3 ∞ ∞ n 3 1 −z n z = = X = X(−1)n . z +3 1 − ( −z ) 3 3n 3 n=0 n=0

This series expansion is valid for |z| < 3; so, in particular, it is valid in the annulus 2 < |z| < 3. To expand 2 , in the annulus 2 < |z| < 3, because 2 < 1, we divide the denominator by z and get z+2 z 2 2 2 1 = = . z +2 2 z − −2 z(1 + z 1 z Expand, using a geometric series, which is valid for |2/z| < 1or2< |z|, and get

∞ ∞ 2 2 1 2 −2 2n+1 = = X( )n = X(−1)n z +2 z 1 − −2 z z zn+1 z n=0 n=0 ∞ 2n = X(−1)n−1 , zn n=1 which is valid for 2 < |z|. Hence, for 2 < |z| < 3,

z 3 2 = − (z + 2)(z +3) z +3 z +2 ∞ ∞ zn 2n = X(−1)n − X(−1)n−1 3n zn n=0 n=1 ∞ ∞ zn 2n = X(−1)n + X(−1)n . 3n zn n=0 n=1

17. First, derive the partial fractions decomposition

z2 +(1− i)z +2 1 2 =1+ − . (z − i)(z +2) z − i z +2

(The ﬁrst step should be to reduce the degree of the numerator by dividing it by the denominator. As in Exercise 13, we handle each term separately, the constant term is to be left alone for now. 92 Chapter 4 Power Series and Laurent Series

In the annulus 1 < |z| < 2, we have 1 < 1 and z < 1. So to expand 1 , factor the z in the z 2 z−i denominator and you’ll get 1 1 1 1 = = , z − i − i z − i z(1 z ) 1 z where i < 1or1< |z|. Apply a geometric series expansion: for 1 < |z|, z ∞ ∞ n 1 1 1 1 i n i = = X = X z − i z 1 − i z z zn+1 z n=0 n=0 ∞ in−1 = X . zn n=1 To expand 2 , in the annulus 1 < |z| < 2, because z < 1, we divide the denominator by 2 and z+2 2 get 2 2 1 = = . z − −z z +2 2(1 + 2 1 2 Expand, using a geometric series, which is valid for |z| < 2, and get

∞ ∞ 2 1 −z zn = = X( )n = X(−1)n . z +2 1 − −z 2 2n 2 n=0 n=0 Hence, for 1 < |z| < 2,

∞ ∞ 1 2 in−1 zn 1+ − =1− X + X(−1)n . z − i z +2 zn 2n n=1 n=0

1 21. The function f(z)= has isolated singularities at z = 1 and z = −i.Ifwe (z − 1)(z + i) √ start at the center z0 = −1, the√ closest singularity is −i and its distance to z0 is 2. Thus√ f(z)is analytic in the disk of radius 2 and center at z0 = −1, which is the annulus |z +1| < 2. Moving outside√ this disk, we encounter the second singularity at z = 1. Thus f(z) is analytic in the annulus 2 < |z +1| < 2, and has a Laurent series representation there. Finally, the function is analytic in the annulus 2 < |z +1| and so has a Laurent expansion there. We now derive the three series expansions. Using a partial fractions decomposition, we have 1 A A f(z)= = − , (z − 1)(z + i) z − 1 z + i 1 − i 1 − | | where A = 2 2 = 2 (1 i). We have, for z +1 < 2, 1 1 1 1 = = − z − 1 −2+(z +1) 2 − z+1 1 2 ∞ 1 z +1 n = − X . 2 2 n=0 √ | | z+1 For z +1 < 2, we have 1−i < 1, and so

1 1 −1 1 = = − − − z+1 z + i (i 1)+(z +1) 1 i 1 1−i ∞ −1 z +1 n = X . 1 − i 1 − i n=0 Section 4.5 Laurent Series 93

√ Thus, for |z +1| < 2, we have 1 A A f(z)= = − (z − 1)(z + i) z − 1 z + i ∞ ∞ A z +1 n A z +1 n = − X + X 2 2 1 − i 1 − i n=0 n=0 ∞ ∞ 1 − i z +1 n 1 z +1 n = − X + X . 4 2 2 1 − i n=0 n=0 √ | | 1−i For 2 < z +1, we have z+1 < 1, and so

1 1 1 1 = = z + i (i − 1) + (z +1) z +1 − 1−i 1 z+1 ∞ ∞ 1 1 − i n (1 − i)n = X = X z +1 z +1 (z +1)n+1 n=0 n=0 ∞ 1 (1 − i)n = X . 1 − i (z +1)n n=1 √ So, if 2 < |z +1| < 2, then 1 A A f(z)= = − (z − 1)(z + i) z − 1 z + i ∞ ∞ 1 − i z +1 n 1 − i (1 − i)n = − X − X . 4 2 2 (z +1)n+1 n=0 n=0 Finally, for 2 < |z +1|, we have 1 1 1 1 = = z − 1 −2+(z +1) z +1 − 2 1 z+1 ∞ ∞ 1 2 n 1 2n = X = X . z +1 z +1 2 (z +1)n n=0 n=1 So, if 2 < |z +1|, then 1 A A f(z)= = − (z − 1)(z + i) z − 1 z + i ∞ ∞ 1 − i 2n 1 − i (1 − i)n = X − X 2 (z +1)n+1 2 (z +1)n+1 n=0 n=0 ∞ ∞ 1 − i 2n 1 (1 − i)n = X − X . 4 (z +1)n 2 (z +1)n n=1 n=1

25. In this problem, the idea is to evaluate the integral by integrating a Laurent series term- by-term. This process is justiﬁed by Theorem 1, which asserts that the Laurent series converges absolutely and uniformly on any closed and bounded subset of its domain of convergence. Since a path is closed and bounded, if the path lies in the domain of convergence of the Laurent series, then the series converges uniformly on the path. Hence, by Corollary 1, Sec. 4.2, the series can be 94 Chapter 4 Power Series and Laurent Series

diﬀerentiated term-by-term. We now present the details of the solution. Using the Maclaurin series of sin z, we have for all z =0,6 ∞ 1 (−1)n sin = X z−(2n+1). z (2n + 1)! n=0 Thus ∞ ∞ 1 (−1)n ! (−1)n Z sin dz = Z X z−(2n+1) dz = X Z z−(2n+1) dz. z (2n + 1)! (2n + 1)! C1(0) C1(0) n=0 n=0 C1(0) We now recall the important integral formula: for any integer m: 2πi if m = −1, Z zm dz = C) 0 otherwise, where C is any positively oriented simple closed path containing 0 (see Example 4, Sec. 3.4). Thus, 2πi if n =0, Z z−(2n+1) dz = 0 otherwise. C1(0) Hence all the terms in the series ∞ (−1)n X Z z−(2n+1) dz (2n + 1)! n=0 C1(0) are 0, except the term that corresponds to n = 0, which is equal to 2πi.So 1 Z sin dz =2πi. C1(0) z

29. We follow the same strategy as in Exercise 25 and use the series expansion from Exercise 5. We have ∞ 1 (−1)n−1 1 ! Z Log 1+ dz = Z X dz z n zn C4(0) C4(0) n=1 ∞ (−1)n−1 1 = X Z dz n zn n=1 C4(0) =2πi,

1 where we have used the fact that n dz =2πi if n = 1 and 0 otherwise. RC4(0) z 33. (a) Consider the power series in (1), where the coefficients are given by (2). Since f(z) and 1 n+1 are analytic in the annulus AR ,R (z0), the path in the integral defining the coefficients (z−z0) 1 2 in (2), CR(z0), can be replaced by any positively oriented path Cr(z0), where R1

reason for this is that Cr(z0) and CR(z0) are mutually deformable in AR1,R2 (z0). Fix z, r1 and

r2, such that z ∈ AR1 ,R2 (z0) is arbitrary but ﬁxed, R1

1 f(ζ) |a | = Z dζ n 2πi (ζ − z )n+1 Cr (z0) 0 2 1 M M ≤ 2πr2 n+1 = n . 2π r2 r2 Section 4.5 Laurent Series 95

z−z0 For r1 < |z − z0|

M − n ≤ | − |n n an(z z0) n z z0 Mρ , r2

∞ n n and so Pn=0 an(z − z0) converges absolutely, by comparison with the series M P ρ . Since z is ∞ − n arbitrary in AR1,R2 (z0), we conclude from Lemma 1, Sec. 4.3, that the series Pn=0 an(z z0) converges absolutely for all |z − z0|

Note that this proof shows that the series in (1) deﬁnes an analytic function in the disk BR2 (z0). Were it not for the other series with negative powers in (1), the function f(z) would be analytic in

BR2 (z0). (b) The proof in this part is very similar to the proof in part (a). In fact, we will use the notation from (a). Fix z, r1 and r2, such that z ∈ AR1 ,R2 (z0) is arbitrary but ﬁxed, R1

1 f(ζ) |a | = Z dζ n 2πi (ζ − z )n+1 Cr (z0) 0 1 1 M M ≤ −n 2πr1 n+1 = n = Mr1 . 2π r1 r1 Equivalently, for n ≥ 1, n |a−n|≤Mr1 .

r1 For r1 < |z − z0|

a Mrn −n ≤ 1 = Mρn, n n (z − z0) |z − z0|

∞ a−n n and so n converges absolutely, by comparison with the series M ρ . Since z is arbi- Pn=0 (z−z0) P ∞ a−n trary in AR ,R (z0), we conclude that the series n converges absolutely for all AR ,R (z0). 1 2 Pn=0 (z−z0) 1 2

To show that the series converges uniformly on closed subsets of AR1,R2 (z0), it suﬃces to show that it converges uniformly on any closed subannulus Ar1 ,r2 (z0), where R1

Solutions to Exercises 4.6 1. The function f(z)=(1− z2) sin z has zeros at ±1 and kπ, where k is an integer. All these zeros are simple zeros. For z = 1, write f(z)=(z − 1)(z + 1) sin z =(z − 1)λ(z). Since λ(1) =0,we6 conclude that z = 1 is a simple zero. A similar argument applies to z = −1. For the remaining zeros, which are the zeros of sin z, see Example 1. 5. We will use the fact that the zeros of sin z are all isolated simple zeros. At z = 0, we have sin z = zλz, where λ0 =6 0. So, sin7 z f(z)= = z3λ7(z) λ7(0) =06 . z4 This shows that if we deﬁne f(0) = 0, then f(z) has a zero of order 3 at 0. All other zeros of f occur at the zeros of sin z, kπ, k =6 0, and they are of order 7. 9. We have z2 z4 z6 cos z =1− + − + ··· 2! 4! 6! So z2 z4 z6 1 z2 f(z)=1− − cos z = − + + ···= z4 − + + ···. 2 4! 6! 4! 6! − 1 z2 ··· Note that λ(z)= 4! + 6! + is a power series that converges for all z. Thus λ(z) is entire. − 1 6 2 6 Moreover, λ(0) = 4 = 0. Thus f(z)=z λ(z), where λ(z) is analytic and λ(0) = 0, implying that f(z) has zero of order 2 at 0. 13. Clearly, the function 1 − z2 z − 1 f(z)= + sin z z +1 has isolated singularities at −1 and kπ, where k is an integer. These singularities are all simple poles. To prove the last assertion, it is easier to work with each part of the function separately. 1−z2 First, show that sin z has a simple pole at the zeros of sin z, which follows immediately from the z−1 − fact that the zeros of sin z are simple zeros. Second, show that z+1 has a simple pole at z = 1, which follows immediately from the fact that −1 is a simple zero of z + 1. Now to put the two terms together, you can use the following fact: If f(z) has a pole of order m at z0 and g(z) is analytic at z0, then f(z)+g(z) has a pole of order m at z0. This result is easy to prove using, for example, Theorem 8. 17. Write 1 1 sin z z tan = z 1 . z cos z 1 The problem points of this function are at 0 and at the zeros of the equation cos z = 0. Solving, we ﬁnd 1 π 2 = + kπ ⇒ z = z = ,kan integer. z 2 k π(2k +1)

Since, as k →∞, zk → 0, the function f(z) is not analytic in any punctured disk of the form 0 < |z|. Thus 0 is not an isolated singularity. At all the other points zk, the singularity is isolated and the order of the singularity is equal to the order of the zero of cos z at zk. Since the zeros of cos z are all simple (this is very similar to Example 1), we conclude that f(z) has simple poles at zk. 21. We have an isolated singularity at 0, which is clearly an essential singularity. To see this, we consider the Laurent series expansion of f(z): 1 1 1 1 1 1 1 1 − sin = − − + −··· = − + ···. z z z z 3!z3 5!z5 3!z3 5!z5 Section 4.6 Zeros and Singularities 97

1 ∞ 25. Determining the type of singularity of f(z)= z+1 at is equivalent to determining the type of singularity of 1 1 z f = 1 = z z +1 1+z 1 at z = 0. Since f z has a removable singularity at 0, we conclude that f has a removable singularity at z = ∞. Note that this is consistent with our characterization of singularities according to the behavior of the function at the point. Since f(z) → 0asz →∞, we conclude that f has a removable singularity and may b redeﬁned to have a zero at ∞. 1 29. Arguing as in Exercise 25, it follows that sin has a removable singularity at ∞ and may be z redeﬁned to have a zero at ∞. 1 33. (a) Suppose that f is entire and bounded. Consider g(z)=f( z ). Then g is analytic at all 6 6 | | | 1 |≤ ∞ z = 0. So z = 0 is an isolated singularity of g(z). For all z = 0, we have g(z) = f( z ) M< , where M is a bound for |f(z)|, which is supposed to exist. Consequently, g(z) is bounded around 0 and so 0 is a removable singularity of g(z). (b) Since f is entire, it has a Maclaurin series that converges for all z. Thus, for all z, f(z)= ∞ n 1 6 sumn=0anz . In articular, we can evaluate this series at z and get, for z =0,

∞ a g(z)=f(z)=X n . zn n=0 By the uniqueness of Laurent series expansion, it follows that this series is the Laurent series of g. But g has a removable singularity at 0. So all the terms with negative powers of z must be zero, implying that g(z)=a0 and hence f(z)=a0 is a constant.

37. (a) If f has a pole of order m ≥ 1atz0, then 1 f(z)= m φ(z), (z − z0) where φ is analytic at z0 and φ(z0) =6 0. (See (6), Sec. 4.6.) So if n is a positive integer, then 1 1 [f(z)]n = φn(z)= ψ(z), mn mn (z − z0) (z − z0)

n where ψ is analytic at z0 and ψ(z0) =6 0. Thus [f(z)] has a pole at z0 of order mn if n>0. If n<0, then 1 [f(z)]n =(z − z )−mn =(z − z )−mnψ(z), 0 φn(z) 0 n where ψ is analytic at z0 and ψ(z0) =6 0. Thus [f(z)] has a zero at z0 of order −mn if n<0. (b) If f has an essential singularity at z0 then |f(z)| is neither bounded nor tends to inﬁnity at z0. n n Clearly, the same holds for |[f(z)] | = |f(z)| : It is neither bounded nor tends to ∞ near z0. Thus n [f(z)] has an essential singularity at z0. 41. Suppose that f and g are analytic in a region Ω and fg is identically zero in Ω. Suppose that g(z) is not identically 0 in Ω and let z0 be such that g(z0) =6 0. Because g is continuous and g(z0) =,6 we can ﬁnd a neighborhood of z0, Br (z0), such that g(z) =6 0 for all z ∈ Br(z0). Since f(z)g(z)=0 for all z ∈ Ω, it follows that f(z) = 0 for all z ∈ Br (z0). Thus, the zeros of f are not isolated, and so, by Theorem 3, f is identically 0. 45. Suppose that p(z) is a polynomial such that |p(z)| = 1 for all |z| = 1. Since |p(z)| = 1 for all |z| = 1, the polynomial has no zeros on the unit circle. Let m be the degree of the polynomial and let a1, a2,. . . ,an denote the zeros of p(z) inside the unit disk, counted according to multiplicity. 98 Chapter 4 Power Series and Laurent Series

z−aj Then n ≤ m. For each j =1, ...,,n, consider the linear fractional transformations φa (z)= . j 1−aj z n We know form Example 3, Sec. 3.7, that |φj(z)| = 1 for all |z| = 1. Let φ(z)=Πj=1φj (z) denote the product of the φj, where each φj is repeated according to the multiplicity of the zero. We have |φ(z)| = 1 for all |z| = 1. Moreover, φ has exactly n zeros at the aj. (If some of the aj ’s are repeated, then the order of the zero at aj is equal to the number of times we repeat aj .) Consider p(z) n 1 − ajz g(z)= = p(z)Πj=1 . φ(z) aj − z

Factoring out the zeros in p(z) and canceling the factors (z − aj) in the numerator with the same in the denominator in g(z), we see that g(z) has removable singularities at aj , hence it can be redeﬁned to be analytic at these points, and so we will consider g(z) to be analytic in B1(0). Moreover, g(z) has | | | | |p(z)| no zeros in B1(0), and for z = 1, we have g(z) = |φ(z)| = 1. By the maximum modulus principle (Corollary 3(ii), Sec. 3.7), it follows that g(z)=A is constant in B1(0), and since |g(z)| = 1 for |z| = 1, it follows that |A| = 1. As a consequence, we have

n aj − z p(z)=AΠj=1 . 1 − aj z Thus n n p(z)Πj=1(1 − ajz)=AΠj=1(aj − z). On the right side we have a polynomial of degree n. On the left side, we have a polynomial p(z) of degree m ≥ n, multiplied by a polynomial of degree equal to the number of nonzero aj ’s. It is clear that such an equality is impossible unless m = n and all aj = 0; otherwise the degree of the polynomial on the left becomes strictly greater than the degree of the polynomial on the right. Consequently, p(z)=A(−)nzn = Bzn, where |B| =1. Section 4.7 Harmonic Functions and Fourier Series 99

Solutions to Exercises 4.7 1. (a) The graph of the boundary function

π + θ if − π ≤ θ ≤ 0, f(θ)= π − θ if 0 <θ<π, is a triangular wave that repeats every 2π-units. See Figure 8, Sec. 7.2. (b) The solution of the Dirichlet problem with boundary function f(θ) is given by

∞ n u(r, θ)=a0 + X r (an cos nθ + bn sin nθ)(0≤ r<1), n=1 where an and bn are the Fourier coeﬃcients of f, given by formulas (6)–(8). However, since the formula for f is given on the interval [−π, π], it is more convenient to use equivalent formulas that are obtained by changing the interval of integration in (6)–(8) from [0, 2π]to[−π, π]. More precisely, we can use π 1 Z a0 = f(θ) dθ, 2π −π π 1 Z an = f(θ) cos nθ dθ (n ≥ 1), π −π π 1 Z bn = f(θ) sin nθ dθ (n ≥ 1). π −π

The reason is that the preceding integrands are 2π-periodic, and so the values of the integrals are the same over any interval of length 2π. (See Theorem 1, Sec. 7.1) (c) Since f is even, we have

π π π 1 Z 2 Z 1 Z π a0 = f(θ) dθ = f(θ) dθ = (π − θ) dθ = . 2π −π 2π 0 π 0 2 Similarly, since f(θ) cos nθ is even,

π π 1 Z 2 Z an = f(θ) cos nθ dθ = (π − θ) cos nθ dθ π −π π 0 2 cos nθ sin nθ π 2 1 cos nπ = − − (θ − π) = − . 2 2 2 π n n 0 π n n Now using the fact that f(θ) sin nθ is odd, we obtain

π 1 Z bn = f(θ) sin nθ dθ =0 (n ≥ 1). π −π (d) Plugging the coeﬃcients into the series, we obtain the solution for 0

∞ π 4 π 4 r2k+1 u(r, θ)= + X rn cos nθ = + X cos(2k +1)θ. 2 πn2 2 π (2k +1)2 n odd k=0 (e) We know that the solution of the Dirichlet problem inside the unit disk approaches the boundary function as r → 1. Hence limr→1 u(r, θ)=f(θ). Now assuming that limr→1 u(r, θ)=u(1,θ), we obtain u(1,θ)=f(θ), or

∞ π 4 π 4 1 f(θ)= + X cos nθ = + X cos(2k +1)θ. 2 πn2 2 π (2k +1)2 n odd k=0 100 Chapter 4 Power Series and Laurent Series

5. (a) Starting with the solution that we derived in Example 1 and using the series in Exercise 4, we obtain

∞ ∞ 200 1 r k 200 1 u(r, θ)=50+ X sin(2k +1)θ =50+ X ρk sin(2k +1)θ, π 2k +1 2 π 2k +1 k=0 k=0 r ≤ where ρ = 2 ,0 ρ<1. So by Exercise 5, 200 1 ρ sin θ ρ sin θ u(r, θ)=50+ tan−1 + tan−1 π 2 1 − ρ cos θ 1+ρ cos θ 100 r sin θ r sin θ ==50+ tan−1 + tan−1 π 2 − r cos θ 2+r cos θ 100 y y =50+ tan−1 + tan−1 , π 2 − x 2+x

where we have used the relations x = r cos θ and y = r sin θ. (b) Let 0

100 −1 y −1 y T =50+ π tan 2−x + tan 2+x , π − −1 y −1 y 100 (T 50) = tan 2−x + tan 2+x .

tan a+tan b Apply the tangent to both sides and use the identity tan(a + b)= 1−tan a tan b . To simplify notation, π − write A = tan 100(T 50). Then

π y + y tan (T − 50) = 2−x 2+x , − y y 100 1 2−x 2+x −4y A = , x2 + y2 +4 4 x2 + y2 +4=−4Ay, x2 + y2 + y = −4, A 2 2 4 x2 + y + = − 4. A A2

This is the equation of a circle centered at (0,y1), where 2 2 π y = − = − = −2 cot (T − 50) , 1 A π − 100 tan 100(T 50) and radius

r 4 r 1 r π R = − 4=2 − 12 cot2 (T − 50) − 1 A2 A2 100 r π π =2csc2 (T − 50) = 2 csc (T − 50) . 100 100

π − Note that the cosecant is positive for 100 (T 50), with 0