Midterm Solutions - Math 44/58, Fall 24. Let f be holomorphic in an open connected set containing the annulus {z C : r z z r 2 }, where < r < r 2. (a Use an appropriate contour to obtain an integral self-reproducing formula analogous to the Cauchy integral formula for f(z in terms of the values of f on C r and C r2. Here C r = {z C : z z = r}. (b Use the formula you obtained in part (a to derive the Laurent series expansion of f: f(z = a n (z z n, n= and verify that it converges absolutely and uniformly on the annulus. (c Derive integral expressions for a n in terms of f analogous to the derivative forms of Cauchy integral formula. z z C r C r2 Proof. Let C r, C r2 denote the circle centered at z radius r, r 2 respectively with < r < r 2. Since C r, C r2 is compact living in the open set Ω (i.e. disjoint from the closed set Ω C there is some δ neighborhood of C r, C r2 that live inside Ω. Let ρ, ρ 2 be such that < ρ < r < r 2 < ρ 2 and C ρ, C ρ2 Ω. Let S z be the boundary of the gray sector depicted in the figure which does not contain z. Let R z be the boundary of the remaining white region. Here all integrals over closed curve are taken counterclockwise which we denote by. Let g(ζ =. Hence by Cauchy s theorem, ζ z g(ζdζ =, g(ζdζ = f(z S z R z
2 Here our integrals are taken in anticlockwise direction. Hence ( f(z = [ ] g(ζdζ + g(ζdζ = [ S Z R Z ζ z dζ C r2 C r ] ζ z dζ Note that M for some M > on C ρ2 or C ρ which are compact. Now if ζ C ρ2 and z z r 2 then z z ζ z < hence ζ z = (ζ z (z z = (ζ z ( z z ζ z = ζ z ( z z n= ζ z We claim that ( z z n ζ z n= ζ z converges uniformly on {z : z z r 2 }. To see this,if z z r 2 we have ( n z z M n r 2 ζ z ζ z r 2 ρ 2 n= The series on RHS is convergent as r 2 /ρ 2 < and is independent of z hence it is easy to see that the convergence is uniform in the region. By uniform convergence (it is also uniform in ζ C ρ2, we may swap the sum and integral: C r2 ζ z dζ = n= ( C r2 In the same fashion, suppose z z r, ζ C ρ ζ z = n= dζ (z z (ζ z n+ n then (ζ z (z z = (z z ( ζ z z z = z z For all such z, we have z z ( n ζ z M n= z z r n= ( n ρ r n ( ζ z n= z z the series on RHS is convergent as ρ /r < as is independent of z hence the series is uniformly convergent on z z ρ. Swapping the sum and the integral, one has C r ζ z dζ = ( (ζ z n dζ (z z (n+ n= C r = ( (ζ z n dζ (z z n C r n= n
By equation (, we have f(z = [ C r2 ζ z dζ ( = = where n= n= C r2 a n (z z n C r dζ (ζ z n a n = ] ζ z dζ (z z n + { C r2 n= (ζ z dζ if n C n+ r (ζ z dζ if n <. n+ ( (ζ z n dζ (z z n C r where the series converges uniformly on {z : r z r 2 } Ω. Also we may obtain a more general formula from homotopy form of Cauchy s theorem. Two paths γ, γ with the same endpoints inside Ω is said to be homotopic if γ can be continuously deformed to γ with their endpoints keep fixed. Cauchy s theorem states that if f is holomorphic in Ω then dζ = γ dζ γ so we can write a n = γ dζ (ζ z n+ for any simple closed curve γ Ω, enclosing z. 2. (a Determine whether the limit R lim R R e ix2 dx exists. If yes, find its value. If not, justify why not. π 4 R
4 Proof. Consider the positively oriented closed contour Γ(R consisting of the three curves Γ (R := [, R], Γ 2 (R := {Re iθ : θ π 4 }, Γ (R := {te i π 4 : t R}, (2 Since the function f(z = e iz2 is entire, Cauchy s theorem gives Γ R f(z dz =, or, I (R + I 2 (R + I (R = for every R. Here I j (R denotes the integral of f over Γ j (R with orientation consistent with Γ. We observe that R I (R = e it2 e i π 2 + i 2 I 2 (R = R π 4 π 4 e i π 4 dt = + i 2 R e t2 dt = e ir2 e 2iθ ire iθ dθ ( + i 2 π 4 e cr2θ dθ e dr2 cr e t2 dt π, as R, whereas 2 e R2 sin(2θ R dθ 2 cr. Here c and d are fixed positive constants. Inserting these estimates into (2, we obtain that R π( + i lim e ix2 dx = 2 lim I (R =. R R R 2 (b By integrating a branch of log z/(z around the boundary of an indented sector of aperture 2π, show that log x 4π2 dx = x 27.
5 ɛ R ( Proof. We choose the contour Γ(ɛ, R to be a positively oriented closed curve, consisting of the following pieces: Γ (R = [, R], Γ 2 (R = {Re iθ : θ 2π }, Γ (ɛ, R = {te : t [ɛ, ɛ] [ + ɛ, R]}, Γ 4 (ɛ = {e + ɛe iθ : π θ 2π }, Γ 5 (ɛ = {ɛe iθ : θ 2π }. Define a complex branch of the logarithm on C \ (, ], and note that for every < ɛ R, the function f(z = log z/(z is holomorphic on an open set containing Γ(ɛ, R and its interior. In particular, observe that z = is a removable singularity for f, and that the only pole of f that Γ(ɛ, R approaches arbitrarily closely is z = e. Since Γ(ɛ, R is homotopic to zero, Cauchy s theorem implies that f(z dz =. Γ(ɛ,R We will show below that (4 lim f(z dz =, R (5 (6 lim ɛ R Γ 2 (R On the other hand, it is easy to verify that f(z dz = e Γ (ɛ,r lim f(z dz =, ɛ Γ 5 (ɛ log x + x dx, lim = πires(f; e ɛ Γ π 2 = 4 (ɛ 9 e ( + i.
6 Assuming these for the moment, the evaluation of the integral is completed as follows. Let I denote the integral to the determined, and let J be the principal value integral given by dx x dx. J = lim ɛ x (, x >ɛ. Letting R and ɛ in (, and using (4, (5 and (6, we obtain that I + e π2 (I + J + 9 e ( + i =. Multiplying both sides of the equation above by e, and then equating real parts of both sides, we obtain that I = 2 π2 4π2, or I =. 2 9 27 Proof of (4. It remains to prove (4. We estimate both integrals by parametrizing the respective curves. 2π f(z dz log(re iθ R e iθ Rieiθ dθ Γ 2 (R Γ 2 (ɛ C R log R as R. R 2π f(z dz log(ɛe iθ ɛ e iθ ɛieiθ dθ ɛ log ɛ C ɛ as ɛ.. Justify the following statements. (a If m and n are positive integers, then the polynomial p(z = + z + z2 2! + + zm m! + zn has exactly n zeros inside the unit disc, counting multiplicities. Proof. Let p(z = + z + z2 + + zm + 2! m! zn then on {z; z = } we have p(z + ( z n = + z + zm m! z m e z < m! Hence for z =, p(z + ( z n < p(z + z n. By Rouche s Theorem, p(z and z n have the same number of zeros in {z : z < } and it is easy to see that z n has n roots (counted with multiplicities with z <, the same is true for p(z. Note: You can t compare two complex numbers, just their absolute values. n=
(b For any λ C with λ < and for n, the function (z n e z λ has n zeros satisfying z < and no other zeros in the right half plane. Proof. Let f(z = (z n e z λ, λ < and g(z = (z n e z. Let Ω = {z : z < } On Ω we have R(z > and f(z. Hence f(z + g(z = λ < (z n e z = g(z < f(z + g(z By Rouche s Theorem, f, g have the same number of zeros inside Ω. Since e z has no zeros, it is easy to see that g has n zeros (counted with multiplicities inside Ω and hence the same is true for f. On {R(z > } { z > }, we have (z n e z = z n e R(z e > λ hence no other zeros on the right half plane. 7 4. Let f be analytic in the punctured disc G = B(a; R \ {a}. (a Show that if f(x + iy 2 dy dx <, then f has a removable singularity at z = a. G (b Suppose that p > and f(x + iy p dy dx <. G What can you conclude about the nature of singularity of f at z = a? Proof. WLOG, assume a =. Suppose f has a pole of order n at then f(z = g(z z n for some g holomorphic on B(, R, g(. For ɛ > small enough we have C 2 > g(z > C > on B(, ɛ for some C, C 2. Now using polar coordinate, B(,ɛ\{} g(x, y p dxdy < z n B(,ɛ\{} 2π ɛ dxdy < z np r np rdrdθ < np + > p < 2 n Hence f has a pole of order n which is the biggest positive integer that is less than 2/p. If f has a essential singularity at then by considering its Laurent s Series, we may find a sequence of functions g N which has a pole of order N at and g N f uniformly on δ z ɛ for any δ >. Hence as N g N p f p δ z ɛ δ z ɛ
8 But when N is large enough so that p 2 then N lim sup g N p = M M z ɛ This would imply lim sup f p = M M z ɛ so f cannot have essential singularity at. In particular if p = 2, we must have a removable singularity there. We conclude that the integral of f p on G cannot converge for any p > if f has an essential singularity at a. Note: We may not have uniform convergence B(, R\{} so we may not directly interchange the limit and the integral directly on this domain. 5. Determine whether each of the following statements is true or false. Provide a proof or a counterexample, as appropriate, in support of your answer. (a There exists a function f that is meromorphic on C such that Res(f; a. a C pole of f Here Res(f; a denotes the residue of f at a. (Hint: By definition, Res(f; = Res( f;, where f(z = z 2 f( z. Proof. The statement is false. By the residue theorem and a change of variable w = /z, we see that for any meromorphic function f on C and a constant R > sufficiently large, Res(f, = z =R z f( 2 z dz = w =R f(w dw = a C a pole of f Res(f, a. Hence the sum of the residues of a meromorphic function on the extended complex plane is always zero. (b The number of zeros of a meromorphic function in C is the same as the number of poles, both counted with multiplicity. Proof. The statement is true. A meromorphic function in C is a rational function P/Q, where P and Q are polynomials with no common root. Let m and n denote the degrees of P and Q respectively. Then P/Q has exactly m zeros and n poles in C, counting multiplicities. If m = n, these account for all zeros and poles of P/Q in C, and the statement has been verified. If m < n, then P/Q has a zero of multiplicity
(7 n m at, whereas if m > n, then P/Q has a pole of order m n at. Thus the total number of zeros in C matches the total number of poles in all cases. (c For any two polynomials P and Q such that deg(p deg(q 2, and Q only has simple roots, the following identity holds: P(a Q (a =. a:q(a= Proof. The statement is true. Since Q has only finitely many zeros, choose R > large enough so that B(; Rcontains all the roots of Q. Further these roots are simple, hence the rational function P/Q has at most simple poles, located at the roots of Q. By the residue theorem, z =R P (z Q(z dz = On one hand, if a is a root of Q, then a:q(a= (8 Res(a; P/Q = lim z a (z a P (z Q(z = lim z a (9 Res(a; P/Q. P (z Q(z Q(a z a = lim z a P (a Q (a. On the other hand, the assumption deg(p deg(q 2 implies that P (z Q(z dz CRdeg(P deg(q+ as R. z =R Combining (7, (8 and (9 yields the desired claim. 9