Solutions by: KARATUĞ OZAN BiRCAN. PROBLEM 1 (20 points): Let D be a region, i.e., an open connected set in

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1 KOÇ UNIVERSITY, SPRING 2014 MATH 401, MIDTERM-1, MARCH 3 Instructor: BURAK OZBAGCI TIME: 75 Minutes Solutions by: KARATUĞ OZAN BiRCAN PROBLEM 1 (20 points): Let D be a region, i.e., an open connected set in C. (a) Suppose tat u(x, y) : D R as vanising partial derivatives u x and u y at every point in D. Sow tat u is constant on D. Solution: Note tat any two points in a region can be connected by a polygonal line containing only orizontal and vertical line segments. Let (a, b) and (c, d) be any two points in D. Ten tere exists a polygonal line connecting (a, b) and (c, d). Since u x and u y vanis at every point, by te mean value teorem, te cange in u between te successive vertices is 0. To be more precise, if (a, b) is connected to (a +, b) by a orizontal line segment, ten u(a +, b) u(a, b) = u x (a + t, b) for some 0 t 1. But since u x = 0 at every point in D, we conclude tat u(a +, b) = u(a, b). Similarly te same olds in te vertical direction. Tus u(a, b) = u(c, d) and ence u is constant on D. (b) Suppose tat f : D C is an analytic function suc tat f (z) = 0 for every z D. Sow tat f is constant on D. Solution: Suppose tat f = u + iv. Since f is identically zero in D, we obtain tat te partial derivatives u x, u y, v x and v y are zero trougout D. Ten by part (a), we conclude tat f is constant on D. 1

2 2 (c) Suppose tat f : D C is an analytic function suc tat at every point z D, eiter f(z) = 0 or f (z) = 0. Sow tat f is constant on D. Solution: Note tat f 2 is analytic on D and (f 2 (z)) = 2f(z)f (z). Since eiter f(z) = 0 or f (z) = 0, we obtain (f 2 (z)) = 0 for every z D. Ten by part (b), we conclude tat f 2 is constant wic implies tat f 2, and ence f is constant. By part (d), te function f is constant on D. (d) Let f : D C be an analytic function. Prove tat if f is constant on D, ten so is f. Solution: If f = 0, ten f = 0. So assume tat f = u + iv = c > 0 for some constant c. Ten u 2 +v 2 = c. Taking te partial derivatives gives us te equations uu x + vv x = 0 and uu y + vv y = 0. Now using te Caucy-Riemann equations, we obtain uu x vu y = 0 and uu y + vu x = 0. Ten (uu x vu y ) 2 = u 2 u 2 x 2uvu x u y + v 2 u 2 y = 0 and (uu y + vu x ) 2 = u 2 u 2 y + 2uvu x u y + v 2 u 2 x = 0. Hence we obtain, (u 2 + v 2 )(u 2 x + u 2 y) = c(u 2 x + u 2 y) = 0 and tis implies tat u x = 0 and u y = 0. Moreover by te Caucy-Riemann equations, v x = v y = 0. Hence f is constant by part (b). PROBLEM 2 (10 points): Let f(z) = 1 z and let (t) = Reit (0 t π). Sow tat lim R f(z)dz = 0. Solution: Note tat f(z)dz f(z) dz. Also by te triangle inequality, we obtain z z 3. Terefore we get z z and tis implies te inequality z z 3 1.

3 Using te M-L inequality, we obtain f(z) dz 1 = z dz 1 z 3 1 dz 1 Combining wit te inequality f(z)dz limit gives te desired result. PROBLEM 3 (15 points) Define sin z = eiz e iz Sow tat for all z, w C, 2i cos(z + w) = cos z cos w sin z sin w. 3 R 3 1 (πr). f(z) dz and taking te and cos z = eiz + e iz. 2 Solution: Using te definitions of te cosine function, cos(z + w) = ei(z+w) + e i(z+w). 2 On te oter and, we ave cos z cos w = eiz + e iz e iw + e iw 2 2 and sin z sin w = eiz e iz e iw e iw 2i 2i Hence cos z cos w sin z sin w = 2ei(z+w) + 2e i(z+w) desired. = ei(z+w) + e i(z w) + e i( z+w) + e i( z w) 4 = ei(z+w) e i(z w) e i( z+w) + e i( z w). 4 4 = ei(z+w) + e i(z+w), as 2 PROBLEM 4 (20 points): (a) Determine te radius of convergence of te power series z 2n f(z) =. n=0

4 4 (b) Sow tat z 2 f (z) + zf (z) = 4z 2 f(z). Solution: We use ratio test. z 2n+2 ((n)!) 2 z 2 lim = 0 n ((n + 1)!) 2 z 2n n (n + 1) 2 and ence te power series converges for all z. Now, we sow tat te series satisfies te differential equation. Note tat f (z) = 2nz 2n 1 and f (z) = (2n)(2n 1)z 2n 2. Hence we obtain z 2 f (z)+zf (z) = (2n)(2n 1)z 2n + 2nz 2n (n!) = (2n)(2n)z 2n 2 = 4z 2n ((n 1)!) 2 = n=0 4z 2n+2 = 4z 2 n=0 z 2n = 4z2 f(z). PROBLEM 5 (20 points): Consider te function f : C C defined by 0 if x = y = 0 f(x + iy) = x 3 y 3 x 2 + y + + y 3 2 ix3 oterwise x 2 + y 2 (a) (2 points) Express f(x + iy) as u(x, y) + iv(x, y). ( Write out te functions u(x, y) and v(x, y) explicitly!) Solution: Note tat and u(x, y) = v(x, y) = 0 if x = y = 0 x 3 y 3 oterwise x 2 + y 2 0 if x = y = 0 x 3 + y 3 oterwise. x 2 + y 2

5 5 (b) (3 points) Sow tat u x (0, 0), u y (0, 0), v x (0, 0), v y (0, 0) exist. Solution: We calculate te following limits. u x (0, 0) u(, 0) u(0, 0) u y (0, 0) u(0, ) u(0, 0) v x (0, 0) v(, 0) v(0, 0) v y (0, 0) v(0, ) v(0, 0) = 1. = 1. = 1. = 1. (c) (3 points) Sow tat f satisfies te Caucy-Riemann equations at te point (x, y) = (0, 0). Solution: In part b, we found tat u x (0, 0) = 1, u y (0, 0) = 1, v x (0, 0) = 1 and v y (0, 0) = 1. Since u x (0, 0) = v y (0, 0) and u y (0, 0) = v x (0, 0), te function f satisfies te Caucy-Riemann equations at te point (0, 0). (d) (4 points) Sow tat f is not differentiable at te origin. Solution: For = w + iw, we obtain f() f(0) lim 2w 3 i 2w 2 w + iw iw w + iw = i 1 + i.

6 6 On te oter and, if is a real number, we obtain f() f(0) + i lim = 1 + i. Tus te limit does not exist and ence f is not differentiable at te origin. (e) (3 points) Is f differentiable at z = i? If so calculate f (i). Solution: u x (0, 1) u(, 1) u(0, 1) ( 2 + 1) = 0. (1 + ) 3 v(0, 1 + ) v(0, 1) (1 + ) 1 v y (0, 1) = 1. Since u x (0, 1) v y (0, 1), te function f is not differentiable at z = i, since it does not satisfy te Caucy-Riemann equations at z = i. PROBLEM 6 (15 points): If z 1, z 2, z 3 are te vertices of an equilateral triangle in C, ten sow tat z z z 2 3 = z 1 z 2 + z 2 z 3 + z 3 z 1. Hint: If we rotate one side of an equilateral triangle by an angle of π/3 in te appropriate direction we obtain anoter side. Solution: Observe tat (z 3 z 2 )e iπ/3 = (z 1 z 2 ) and (z 3 z 1 )e iπ/3 = (z 3 z 2 ). Ten we get z 1 z 2 = z 3 z 2 z 3 z 2 z 3 z 1 wic gives te desired result after cross multiplication.