Math 185 Fall 2015, Sample Final Exam Solutions

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1 Math 185 Fall 2015, Sample Final Exam Solutions Nikhil Srivastava December 12, True or false: (a) If f is analytic in the annulus A = {z : 1 < z < 2} then there exist functions g and h such that g is analytic in z < 2, h is analytic in z > 1, and f = g + h in A TRUE. We can write f as a Laurent series about zero: f(z) = a n z n + n=0 where both series are convergent in A. Take g(z) = n=0 a nz n and h(z) = b n m=1 ; since the region of convergence of g(z) is a disk centered at zero (because z n it is a power series), g(z) must be convergent in a disk containing A, so it must be convergent in D(0, 2). By the converse to Taylor s theorem, it is analytic there. Similarly, h(z) is convergent in the complement of a closed disk centered at zero (because it is a power series in 1/z) containing A. The smallest such region is z > 1, and by uniform convergence h(z) is analytic there. m=1 b n z n, (b) There is a simple closed contour in C \ {0} such that 1 dz = 2πi z2 C. FALSE. Since 1/z 2 has an antiderivative in C \ {0} every closed integral is zero. Alternatively, by the residue theorem, it has only one singularity (at z 0 = 0) with residue zero. (c) If f(z) and g(z) have a pole at z 0 then Res(fg, z 0 ) = Res(f, z 0 )Res(g, z 0 ). FALSE. For instance f(z) = 1/z and g(z) = 1 + 1/z 2 have a pole at zero with Res(f, 0) = 1, Res(g, 0) = 0, but the product f(z)g(z) = 1/z + 1/z 3 has residue 1 at zero. (d) Every function analytic in D(0, 1) has an analytic continuation to D(0, 2). 1

2 FALSE. The function f(z) = 1 is analytic in D(0, 1) but there is no F (z) 1 z analytic in D(0, 2) with F (z) = f(z) in D(0, 1). The reason is that for a sequence, such as z n = 1 1/n, approaching the pole z 0 = 1 we have so F cannot be analytic there. lim F (z) = lim f(z) =, n n (e) If u(x, y) and v(x, y) are harmonic in a domain D then the product u(x, y)v(x, y) is also harmonic in D. FALSE. Consider u(x, y) = v(x, y) = x, clearly harmonic in R 2, but (uv)(x, y) = x 2 which has u xx + u yy = (i) Find a Möbius transformation T mapping the points z 0 = 0, z 1 = i+e iπ/4, z 2 = i+1 to 0, 1, respectively. (ii) Let D be the intersection of the two open disks D(1, 1) = {z : z 1 < 1} D(i, 1) = {z : z i < 1}. Verify that T maps D to the quadrant Q = {z : Re(z) > 0, Im(z) > 0}. Solution: (i) T is given by: T (z) = z 0 i + e iπ/4 (i + 1) z (i + 1) i + e π/4 0 = z e iπ/4 1 z (i + 1) e iπ/4 + i. (ii) Observe that the two circles whose arcs form the boundary of D, namely C 1 : { z 1 = 1} and C 2 : { z i = 1}, intersect at two points: z 0 = 0 and z 2 = i + 1. Since T (z 0 ) = 0 and T (z 2 ) = the images of these circles are lines L 1 and L 2 through the origin. Since z 2 C 2 and T (z 2 ) = 1 R, the image of C 2 must be the real line, with D C 2 mapped to the ray (0, ). Notice that the tangents to the circles C 1 and C 2, oriented away from the origin, are perpendicular at zero. In particular, the tangent to C 1 at zero makes a positive angle of π/2 with the tangent to C 2 at zero. Since T is conformal at zero (it is conformal at each point that is not a pole), it preserves angles and we conclude that T (C 1 D) makes an angle of π/2 with T (C 2 D). Thus, we must have T (C 1 D) = {iy : y > 0}, the positive imaginary axis, and we conclude that D gets mapped to Q. 3. Verify that u(x, y) = x 3 y xy 3 is harmonic in R 2. Find a harmonic conjugate v(x, y) of u and an entire function f(z) such that f(x + iy) = u(x, y) + iv(x, y). Write f as a function of z. Solution: We have u xx = d ( 3x 2 y y 3) = 6xy. dx u yy = d ( x 3 3xy 2) = 6xy, dy 2

3 so u xx + u yy = 0 for all (x, y), so u is harmonic everywhere. To find a harmonic conjugate v we can solve the Cauchy-Riemann equations (we know that such a conjugate must exist by HW13, since R 2 is simply connected): v y = u x = 3x 2 y y 3 v(x, y) = (3x 2 y y 3 )dy = 3x 2 y2 2 y4 4 + g 1(x). v x = (u y ) = x 3 + 3xy 2 v(x, y) = (3xy 2 x 3 )dx = 3y 2 x2 2 x4 4 + g 2(y). Solving for g 1 and g 2 we find that v(x, y) = 3 2 x2 y 2 x4 4 y4 4 is a harmonic conjugate of u(x, y). Thus, we have f(x + iy) = (x 3 y yx 3 ) + i( 3 2 x2 y 2 x4 4 y4 4 ). Before we try to figure out what f(z) is, we note that it must be a polynomial of degree 4 in z. For convenience we rewrite: 4f(x + iy) = 4(x 3 y yx 3 ) + i(6x 2 y 2 x 4 y 4 ). and recognize these as essentially (upto a factor of i) the binomial coefficients in the expansion: ( ) ( ) ( ) z 4 = (x+iy) 4 = x 4 + x 3 (iy)+ x 2 (iy) 2 + x(iy) 3 +y 4 = x 4 +y 4 6x 2 y 2 +i(4x 3 y 4xy 3 ) Thus, we have 4f(x + iy) = i(x + iy) 4 and f(z) = iz 4 /4. 4. Consider the polynomial f(z) = z 4 + 5z + 1. How many zeros does f have in the annulus 1 < z < 2? Solution: Notice that when z = 1 we have f(z) > 5z z = 5 2 > 0 so f(z) 0 on z = 1 and 5z > z there. By Rouche s theorem, the number of zeros of f in z < 1 is equal to the number of zeros of 5z there, which is one. On the other hand, when z = 2 we have f(z) > z 4 5z + 1 = > 0, so f(z) 0 on z = 1 and z 4 > 5z + 1 there. By Rouche f must have th same number of zeros in z < 2 as z 4, which is four. Subtracting the two, we conclude that f has exactly 4 1 = 3 zeros in the prescribed annulus. 3

4 5. Prove that there is no entire function with Re(f(z)) = z 2. Solution: If such a function existed, its real part would be harmonic, but the function u(x, y) = x 2 + y 2 has u xx + u yy = , so this is impossible. 6. (i) Find the Taylor expansion of Log(z), the principal branch of the logarithm, centered at z 0 = 2. (ii) Consider the branch Log θ (z) = ln z + iarg θ (z) Arg θ (θ, θ + 2π), for some θ (0, π). What is the radius of convergence of the Taylor series of Log θ (z) at z 0 = 2? Solution: (i) We have: Log(z) = Log(2 + (z 2)) = Log (2 (1 + (z 2)/2)) = Log(1 + z 2 2 ) + ln(2) since multiplication by a real number only changes the magnitude = Log(2) + z 2 2 (z 2) Alternatively, letting f(z) = Log(z), we may compute the derivatives f (z) = 1 z f (z) = 1 z 2 f (z) = 2 z 3 f (n) (z) = ( 1)n 1 (n 1)! z n, so the Taylor coefficients at z 0 = 2 are: a n = f (n) (2) n! = ( 1)n 1 n 2 n, yielding the Taylor series Log(z) = Log(2) (z 2) (z 2 2) (ii) The branch Log θ has a branch cut along the ray R θ : z = re iθ, so the radius of convergence of the Taylor series of Log θ (z) at z 0 = 2 is the radius of the largest open disk centered at z 0 disjoint from R θ. If θ (π/2, π) the ray lies in the left half plane {Re(z) < 0} and the radius of convergence is the distance to zero, which is 2. 4

5 Otherwise, assume θ (0, π/2) and let z be the point on R θ closest to z 0 and note that the segment [z 0, z ] is perpendicular to R θ. Thus, by considering the triangle with vertices 0, z 0, z we have so the radius of the circle is 2 sin(θ). 7. Does the series sin(θ) = z 0 z z 0 0 n=0 z n n! converge uniformly in C? Justify your answer. = z 0 z, 2 Solution: No, this series does not converge uniformly. Consider a partial sum S N = N z n n=0 and let z > 0 be a positive real number. The remainder term is then: n! ρ N (z) = S(z) S N (z) = N+1 z n n! zn+1 (N + 1)!. Taking z = (N + 1)! we have ρ N (z) 1. Thus, for every N there is a z C such that ρ N (z) 1, so the series cannot converge uniformly. 8. Prove the Casorati-Weierstrass theorem: if f(z) has an essential singularity at z 0 then for every w 0, ɛ > 0, and δ > 0, there exists z z 0 < δ such that f(z) w 0 < ɛ. (you can use any theorem on classification of singularities.) Solution: See page 257 of the textbook. 9. Evaluate the integral sin(x) 0 x(x 2 + 1) dx, where we understand the integrand as extending to a continuous function with value lim x 0 sin(x)/(x(x 2 + 1)) at x = 0. Solution: Since the integrand is even, we may write 2I = p.v. where we have taken sin(x) dx = Im x(x 2 + 1) f(z) := dz = Im z(z 2 + 1) lim z(z 2 + 1) dz. ρ 0,R ρ R R f(z)dz+ f(z)dz, ρ The integrand has simple poles at z = 0 and ±i, so we use an indented contour which avoids the pole at zero. Let C 1 : z(t) = z, t [ R, ρ] and C 2 : z(t) = z, t [ρ, R] be contours along the intervals above, and consider the semicircular contours: C ρ : z(t) = ρe it, t [0, π] 5 C R : z(t) = Re it, t [0, π].

6 We then have for every ρ, R > 0: Im f(z)dz = 2I Im C 1 C ρ+c 2 +C R f(z)dz + Im C ρ f(z)dz. C R We evaluate the integral on the left hand side using the residue theorem. For small ρ and large R, the only singularity of f(z) = z(z + i)(z i) inside C = C 1 C ρ + C 2 + C R is at z = i. The residue there is: so Res(f, i) = lim z i z(z + i) = e 1 2, Im f(z)dz = Im(2πi e 1 C 2 ) = π e. On the right hand side, we observe that since 1 in the upper half plane, an ML estimate tells us that e CR iz z(z 2 + 1) dz πr 1 R(R 2 1) 0 as R. The other integral picks up iπ times the residue at zero (since it is oriented clockwise) as ρ 0; this residue is Res(f, 0) = lim z 0 z = 1, so we have Combining these results, we have: Im lim f(z)dz = Im( iπ 1) = π. ρ 0 C ρ π e = 2I π I = π(1 e 1 )/2. 6