COMPLEX ANALYSIS HW 4 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.