127 Consider a Meromorphic Function F : C → C with Laurent Series F(Z) = P∞ N N=−∞ Cnz at the Origin

CLAY SHONKWILER

127 Consider a meromorphic function f : C → C with Laurent series f(z) = P∞ n n=−∞ cnz at the origin. Find a necessary and suﬃcient condition on the complex coeﬃcients (cn) so that f(R) ⊆ R. Claim: If f : C → C is a meromorphic function with Laurent series P∞ n f(z) = n=−∞ cnz at the origin, then f(R) ⊆ R if and only if cn ∈ R for all n ∈ Z.

Proof. If cn ∈ R for all n, then certainly f(R) ⊆ R, so we need only prove that the coefficients being in R is a necessary condition for f(R) ⊆ R. To that end, suppose f(R) ⊆ R. Since f is meromorphic, there can only be finitely many negative terms in the Laurent expansion, so there exists N ∈ N P∞ n such that f(z) = n=−N cnz and c−N 6= 0. Thus, we can re-write f as ∞ ∞ X 1 X f(z) = c zn = c zn. n zN n−N n=−N n=0 P∞ n N Let g(z) = n=0 cn−N z . Then z f(z) = g(z). Now, if r ∈ R, then N r ∈ R and f(r) ∈ R; since the product of two reals is real, this implies that g(r) ∈ R. Since our choice of r was arbitrary, we see that g(R) ⊆ R. To deal with this situation, let us first prove a small lemma: 0 Lemma 0.1. If h ∈ H(C) such that h(R) ⊆ R, then h (R) ⊆ R. Proof. By definition of the derivative, h(z + ζ) − h(z) h0(z) = lim ζ→0 ζ for any z ∈ C. Now, if z ∈ R and ζ = k ∈ R, then h(z+k) ∈ R and h(z) ∈ R, so h(z + k) − h(z) ∈ R. Therefore, since the derivative is well-defined,

0 h(z + k) − h(z) h (z) = lim ∈ R. k→0 k 0 Since our choice of z ∈ R was arbitrary, we conclude that h (R) ⊆ R. First, note that ∞ X n g(0) = cn−N 0 = c−N ∈ R. n=0 1 2 CLAY SHONKWILER

Now, since g is holomorphic and g(R) ⊆ R, we can apply the above lemma 0 to conclude that g (R) ⊆ R. Now, ∞ 0 X n−1 g (z) = ncn−N z , n=1 so ∞ 0 X n−1 g (0) = ncn−N 0 = c1−N ∈ R. n=1 0 00 g also satisfies the hypotheses of the lemma, so g (R) ∈ R, and so ∞ 00 X n−2 g (0) = n(n − 1)cn−N 0 = 2c2−N ∈ R n=2 so c2−N ∈ R. Iterating this process, we see that at the jth stage, ∞ (j) X n−j g (0) = n(n − 1) ··· (n − (j + 1))cn−N 0 = j(j − 1) ··· 2 · 1cj−N ∈ R n=j and so cj−N ∈ R. Therefore, we conclude that cn ∈ R for all n. 2 Let ζ be a point in the unit disk and f ∈ H(D\{ζ}). Show that for all sufficiently small > 0 Z Z f(z)dz = f(z)dz. |z|=1 |z−ζ|= Be careful, use only results that have been established in this course! If E = {ζ1, . . . , ζN } are points in D, and f ∈ H(D\E), then show that, for sufficiently small > 0, N Z X Z f(z)dz = f(z)dz. |z|=1 j=1 |z−ζj |=

Proof. Since f is holomorphic on the unit disk except at ζ, if ρ = min|z|=1{|ζ− z|}, then f has a Laurent expansion in Dρ(ζ)\{ζ} given by ∞ X n cn(z − ζ) . n=−∞

Furthermore, this converges uniformly on any compact subset of Dρ(ζ)\{ζ}. In fact, as |z − ζ| gets larger, the sum −1 X n (1) g(z) = cn(z − ζ) n=−∞ converges more and more rapidly, so in fact g ∈ H(C\{ζ}). Now, let h(z) = f(z) − g(z). COMPLEX ANALYSIS HW 4 3

Then, since f ∈ (D\{ζ}) and g ∈ H(C\{ζ}), h is holomorphic on D\{ζ}. Now, ζ is a removable singularity of h, since in Dρ(ζ)\{ζ} ∞ X n f(z) = cn(z − ζ) + g(z) n=0 and so ∞ X n h(z) = f(z) − g(z) = cn(z − ζ) n=0 in this neighborhood, meaning we can analytically continue h at ζ using this power series. In doing so, we now have h ∈ H(D); hence, by Cauchy- Goursat, Z h(z)dz = 0 γ for any closed curve γ ∈ D. Since f(z) = g(z) + h(z) away from ζ, Z Z Z Z f(z)dz = g(z)dz + h(z)dz = g(z) γ γ γ γ for any γ that doesn’t pass through ζ. Since the sum in (1) converges uniformly outside any small disc centered at ζ, it converges uniformly on γ and so Z −1 Z ∞ Z X X c−n g(z)dz = c (z − ζ)ndz = dz. n (z − ζ)n γ n=−∞ γ n=1 γ Now, if n > 1, then 1 d −1 = (z − ζ)n dz (n − 1)(z − ζ)n−1 so, by the result proved in problem 1 of homework 2, Z c−n −c−n −c−n n dz = n−1 − n−1 = 0 γ (z − ζ) (n − 1)(γ(1) − ζ) (n − 1)(γ(0) − ζ) since γ is a closed curve and so γ(1) = γ(0). Therefore, Z Z Z Z c−1 1 f(z)dz = g(z)dz = dz = c−1 dz. γ γ γ z − ζ γ z − ζ Since this is true for any closed γ ∈ D\{ζ}, the above result holds for γ denoting the circle |z| = 1 − δ for any δ > 0, and so we see that Z Z 1 (2) f(z)dz = c−1 dz. |z|=1 |z|=1 z − ζ Now, Z 1 dz = 2πi |z|=1 z − ζ by Cauchy’s Integral Formula, since the function 1 is holomorphic in D. 4 CLAY SHONKWILER

1 R Now, recall that c−1 is deﬁned to be 2πi |z−ζ|= f(z)dz for 0 < < ρ. Hence, plugging this data into (2), we see that Z 1 Z Z f(z)dz = f(z)dz2πi = f(z)dz. |z|=1 2πi |z−ζ|= |z−ζ|=

To prove the case where f ∈ H(D\E) where E = {ζ1, . . . , ζN }, let

ρ0 = min{|ζ1 − z|,..., |ζN − z|}, |z|=1 let ρi = min{|ζi − ζj|} j6=i and let ρ = min {ρi}. 0≤i≤N Then for each i = 1,...,N, f is holomorphic on the punctured disc of radius ρ centered at ζi and so, for each i, ∞ X ni f(z) = cni z ni=−∞ on Dρ(ζi)\{ζi}. Let −1 X ni (3) gi(z) = cni z n=−∞ and let N X g(z) = gi(z) i=1 Then, by the same argument for the g deﬁned above, g is holomorphic on C\E and so h(z) = f(z) − g(z) is holomorphic on D\E. Now, in Dρ(ζi), ∞ X ni h(z) = f(z) − g(z) = cni (z − ζi) , ni=0 so each ζi is a removable singularity of h, which we can, thus, extend to a holomorphic function on all of D. Thus, by Cauchy-Goursat, Z h(z)dz = 0 γ for any closed curve γ in D. Since f(z) = g(z) + h(z) away from ζi, Z Z Z Z f(z)dz = g(z)dz + h(z)dz = g(z) γ γ γ γ COMPLEX ANALYSIS HW 4 5 P for any γ that doesn’t pass through any of the ζi. Since g = gi is a ﬁnite sum, we can swap sum and integral to see that N Z X Z (4) f(z)dz = gi(z). γ i=1 γ

Furthermore, since γ does not pass through any of the ζi, there is some δ > 0 such that Dδ(ζi) does not intersect γ for all i; since the sum in (3) converges uniformly outside Dδ(ζi), at ζ, it converges uniformly on γ and so Z −1 Z ∞ Z c X ni X −ni gi(z)dz = cni (z − ζ) dz = n dz. γ γ γ (z − ζ) i ni=−∞ ni=1

Now, if ni > 1, then 1 d −1 n = n −1 (z − ζ) i dz (ni − 1)(z − ζ) i so, by the result proved in problem 1 of homework 2, Z c−ni −c−ni −c−ni n dz = n −1 − n −1 = 0 γ (z − ζ) i (ni − 1)(γ(1) − ζ) i (ni − 1)(γ(0) − ζ) i since γ is a closed curve and so γ(1) = γ(0). Therefore, Z Z Z c−1i 1 gi(z)dz = dz = c−1i dz. γ γ z − ζ γ z − ζ Since this is true for any closed γ ∈ D\E, the above result holds for γ denoting the circle |z| = 1 − δ for any ρ > δ > 0, and so we see that Z Z 1 (5) gi(z)dz = c−1i dz. |z|=1 |z|=1 z − ζ Now, Z 1 dz = 2πi |z|=1 z − ζi by Cauchy’s Integral Formula, since the function 1 is holomorphic in D. 1 R Now, recall that c−1 is deﬁned to be f(z)dz for 0 < < ρ. i 2πi |z−ζi|= Hence, plugging this data into (5), we see that Z 1 Z Z gi(z)dz = f(z)dz2πi = f(z)dz. |z|=1 2πi |z−ζi|= |z−ζi|= Therefore, by (4),

N N Z X Z X Z f(z)dz = gi(z)dz = f(z)dz |z|=1 i=1 |z|=1 i=1 |z−ζi|= for any < ρ. 6 CLAY SHONKWILER

3 Compute the following integrals Z z+ 1 e( z )dz |z|=1 Z dz 2 |z|=2 1 + z

z+1/z Answer: The Laurent series for f(z) = e in C\{0} is of the form ∞ X n cnz n=−∞ for Z 1 −(n+1) cn = f(z)z dz 2πi |z|=ρ for ρ > 0 and all n. Specifically, if we let ρ = 1, Z c−1 = f(z)dz. |z|=1 By the uniqueness of the Laurent expansion, if we can find a series of the above form that converges locally uniformly to f, then the coefficient on 1 R z must be |z|=1 f(z)dz. To that end, note that ∞ X wn ew = n! n=0 for any w ∈ C and that this series converges uniformly. Hence, for any z 6= 0, ∞ ∞ n ! ∞ n ! X (z + 1/z)n X 1 X n zk X 1 X n ez+1/z = = = z2k−n n! n! k zn−k n! k n=0 n=0 k=0 n=0 k=0 which converges locally uniformly on C\{0}. Hence, this is the Laurent z+1/z 1 expansion of e , so we need to determine the coefficient on z . It’s clear that, since the power on z is 2k − n, only odd n’s will contribute to this coefficient; specifically, for each odd n, we will pick up a 1 n 1 n! 1 n−1 = n−1 n+1 = n−1 n+1 n! 2 n! 2 ! 2 ! 2 ! 2 ! 1 Therefore, the coefficient on z is ∞ ∞ X 1 X 1 = . (2n+1)−1 (2n+1)+1 n!(n + 1)! n=0 2 ! 2 ! n=0 COMPLEX ANALYSIS HW 4 7

Hence, we conclude that ∞ Z X 1 ez+1/zdz = . (n + 1)!n! |z|=1 n=0 2 1 Turning to the second integral, note that 1+z = (z −i)(z +i), so 1+z2 is holomorphic on D2(0) except at ±i. The result proved in problem 2 above did not depend on the size of the disc, so that result tells us that Z dz Z dz Z dz 2 = 2 + 2 . |z|=2 1 + z |z−i|=1/2 1 + z |z+i|=1/2 1 + z 1 Now, on D1/2(i), the function g(z) = z+i is holomorphic, so, by Cauchy’s Integral Formula, 1 1 Z g(z) 1 Z 1 1 Z dz = g(i) = dz = dz = 2 , 2i 2πi |z−i|=1/2 z − i πi |z−i|=1/2 (z + i)(z − i) 2πi |z−i|=1/2 1 + z R dz 2πi so |z−i|=1/2 1+z2 = 2i = π. 1 On the other hand, h(z) = z−i is holomorphic on D1/2(−i), so Cauchy’s Integral Formula tells us 1 1 Z h(z) 1 Z dz = h(−i) = dz = 2 , −2i 2πi |z+i|=1/2 z + i 2πi |z+i|=1/2 1 + z R dz 2πi so |z+i|=1/2 1+z2 = −2i = −π. Hence, Z dz Z dz Z dz 2 = 2 + 2 = π − π = 0. |z|=2 1 + z |z−i|=1/2 1 + z |z+i|=1/2 1 + z ♣

4 1 The function f(z) = (z−2)(z−1) is meromorphic. Locate its poles. This function can be represented by three diﬀerent Laurent series centered on z = 0. Find them and give their domains of convergence. Answer: f is certainly holomorphic except at z = 2 and z = 1, which are the poles of f. Note that −1 1 1 1 2 −1 (6) = − = z − . (z − 2)(z − 1) z − 2 z − 1 1 − 2 1 − z

−1 2 Now, 1−z/2 is holomorphic on D2(0) and has power series given by the geometric series: ∞ ∞ −1 X z X zn = − . 2 2 2n+1 n=0 n=0 8 CLAY SHONKWILER

−1 On the other hand, 1−z is holomorphic on D1(0) and has power series given by the geometric series: ∞ X − zn. n=0 Hence, applying these power series to (6), we see that ∞ ∞ ∞ ∞ X zn X X zn X 2n+1 − 1 f(z) = − + zn = zn − = zn, 2n+1 2n+1 2n+1 n=0 n=0 n=0 n=0 which is a Laurent (actually Taylor) series expansion of f in D1(0). Now, the Laurent expansion of f in A(0, 1, 2), i.e. the annulus given by 1 < |z| < 2 is deﬁned to be ∞ X n f(z) = cnz n=−∞ where Z Z −(n+1) 1 −(n+1) 1 z cn = f(z)z = 2πi |z|=ρ 2πi |z|=ρ (z − 2)(z − 1) and 1 < ρ < 2. Now, since the result proved in problem 2 above did not depend on the size of the circle, and the poles of f are at 1 and 2, we can use that result to see that (7) 1 Z z−(n+1) 1 Z z−(n+1) 1 Z z−(n+1) cn = = + 2πi |z|=ρ (z − 2)(z − 1) 2πi |z|=1/4 (z − 2)(z − 1) 2πi |z−1|=1/4 (z − 2)(z − 1) if n ≥ 0 and (since then z−(n+1) has a pole at 0) 1 Z z−(n+1) 1 Z z−(n+1) (8) cn = = 2πi |z|=ρ (z − 2)(z − 1) 2πi |z−1|=1/4 (z − 2)(z − 1) otherwise. Now, by the Cauchy Formula, −(n+1) (1)−(n+1) 1 Z z −1 = = z−2 dz 1 − 2 2πi |z−1|=1/4 z − 1 1 Z z−(n+1) = dz. 2πi |z−1|=1/4 (z − 2)(z − 1) On the other hand, if n ≥ 0, then −(n+1) 1 1 Z z 1 Z (z−2)(z−1) dz = n+1 dz 2πi |z|=1/4 (z − 2)(z − 1) 2πi |z|=1/4 z 1 is just the nth coeﬃcient in the power series expansion of (z−2)(z−1) , which 2n+1−1 we calculated above to be 2n+1 . Hence, if n ≥ 0, (7) tells us that 2n+1 − 1 −1 c = − 1 = n 2n+1 2n+1 COMPLEX ANALYSIS HW 4 9 and if n < 0, then (8) tells us that

cn = −1.

Hence, −1 ∞ X X −1 f(z) = −zn + zn, 2n+1 n=−∞ n=0 for z ∈ A(0, 1, 2). Finally, if we consider the annulus A(0, 2, ∞) then, on this annulus, f has a Laurent expansion as before except with ρ > 2. Hence, if n ≥ 0, (9) 1 Z z−(n+1) 1 Z z−(n+1) 1 Z z−(n+1) cn = dz+ dz+ dz 2πi |z|=1/4 (z − 2)(z − 1) 2πi |z−1|=1/4 (z − 2)(z − 1) 2πi |z−2|=1/4 (z − 2)(z − 1) and 1 Z z−(n+1) 1 Z z−(n+1) (10) cn = dz+ dz 2πi |z−1|=1/4 (z − 2)(z − 1) 2πi |z−2|=1/4 (z − 2)(z − 1) otherwise. Now, we’ve already calculated all these terms except

−(n+1) 2−(n+1) 1 Z z 2−(n+1) = = z−1 dz (2 − 1) 2πi |z−2|=1/4 z − 2 1 Z z−(n+1) = dz, 2πi |z−2|=1/4 (z − 2)(z − 1) Hence, if n ≥ 0,

2n+1 − 1 −1 1 c = − 1 + 2−(n+1) = + = 0. n 2n+1 2n+1 2n+1 If n < 0, then −(n+1) cn = −1 + 2 . Hence, plugging into (9) and (10), we see that

−1 ∞ 1 X X −1 + n−1 f(z) = (−1 + 2−(n+1))zn = 2 zn n=−∞ n=1 outside the circle of radius 2. ♣ . 10 CLAY SHONKWILER

5 c Let f ∈ H(Dr(0) ), that is f is holomorphic outside of the disk of radius r. 1 The function F (z) = f( z ) is then holomorphic in a punctured disk, centered at zero. We say that: (a): f has a pole at infinity, if F has a pole at zero. (b): f has an essential singularity at infinity, if F has an essential singularity at zero. (c): f is analytic at infinity, if F has a removable singularity at zero. Prove that a nonconstant, entire function either has a pole or an essential singularity at infinity. Give examples of entire functions with each of these types of singularities. Give an example of a function with a removable singularity at infinity.

Proof. Suppose f is a non-constant entire function. Clearly, f cannot be bounded, else, by Liouville’s Theorem, f would be constant. Furthermore, since f is entire, it is continuous on all of C and so is bounded on any compact subset of C. Since any closed disc centered at the origin is compact, we see that, for |z| ≤ N, |f(z)| is bounded, so |f(z)| is unbounded as |z| → ∞. This leaves two possibilities: either |f(z)| → ∞ as |z| → ∞, or |f(z)| has no limit as |z| → ∞. 1 In the former case, this implies that F (z) = f( z ) is such that |F (z)| → ∞ as z → 0, meaning that F has a pole at zero and so f has a pole at infinity. 1 In the latter case, we see that F (z) = f( z ) is unbounded as z → 0 and yet |F (z)| 6→ ∞ as z → 0, meaning that F is not meromorphic at 0. Since having an essential singularity is equivalent to not being meromorphic, this in turn means that F has an essential singularity at 0, so f has an essential singularity at infinity. Examples: First, consider f(z) = z2. Then f is its own power series and thus is certainly entire. Now, for |z| > 1, |z|2 > |z|, so |f(z)| = |z2| = |z|2 > |z|. Hence, |f(z)| → ∞ as |z| → ∞, which is equivalent to saying 1 that |F (z)| → ∞ as z → 0 for F (z) = f( z ), so F has a pole at 0 and so f has a pole at infinity. Now, consider f(z) = ez. Note first that f is an entire function. Then, if r r r r ∈ R such that r > 0, then e > r, so |f(r)| = |e | = e → +∞ as r → +∞. Hence, f is certainly unbounded. However, 1 e−r = → 0 er as r → +∞, so we see that |f(z)| cannot have a limit at infinity. Hence, f does not have a pole at infinity and so, by the above argument, must have an essential singularity at infinity. 1 1 Finally, consider f(z) = z sin z . Since sin z is entire, sin z ∈ H(C\{0}) 1 sin z and so, since z is entire, f ∈ H(C\{0}). Now, consider F (z) = f z = z , COMPLEX ANALYSIS HW 4 11 which is holomorphic on C\{0}. Now, for z 6= 0, sin z zF (z) = z = sin z → 0 z as z → 0 (since sin 0 = 0), so, by Riemann’s Removable Singularities The- orem, F has a removable singularity at 0, meaning that f has a removable singularity at infinity. ♣

6 f Show that if f ∈ H(D0\{0}) has a nontrivial pole at 0, then e has an essential singularity at 0. A pole is nontrivial if it is not a removable singularity, i.e. some coeﬃcient of a negative degree term in the Laurent expansion is non-zero. Proof. Suppose f has a nontrivial pole at 0. Then f has a Laurent series ∞ X n f(z) = cnz n=−N ∗ where N > 0, c−N 6= 0 and this series converges in some punctured disc D about the origin. Now, g(z) = ez has a uniformly convergent power series ∞ X zn ez = n! n=0 Hence, for any z ∈ D∗, ∞ X (f(z))n ef(z) = n! n=0 where this series converges uniformly for all z 6= 0. Hence, n ∞ P∞ c zj X j=−N j ef(z) = n! n=0 in D∗. If > 0, then for any z ∈ D∗ such that |z| > , both sums converge uniformly, so we can swap the sums: ∞ ∞ X X 1 ef(z) = cnznj. n! j j=−N n=0 Since our choice of was arbitrary, we see that this is a Laurent series for ef . If we restrict our attention to the term given by j = −N, note that n(−N) → −∞ as n → ∞, so we see that there are terms in this Laurent expansion of arbitrarily negative orders. By the uniqueness of the Laurent expansion, this is the Laurent expansion of ef centered at 0, so we see that ef has inﬁnitely many negative terms in its Laurent expansion, which is f precisely what it means for e to have an essential singularity at 0. 12 CLAY SHONKWILER

DRL 3E3A, University of Pennsylvania E-mail address: [email protected]