MTH 3102 Complex Variables Solutions: Practice Exam 2 Mar. 26, 2017

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1 Name Last name, First name): MTH 31 omplex Variables Solutions: Practice Exam Mar. 6, 17 Exam Instructions: You have 1 hour & 1 minutes to complete the exam. There are a total of 7 problems. You must show your work. Partial credit may be given even for incomplete problems as long as you show your work. 1. For the function f z) = exp z) and contour, the boundary of the square with vertices at the points, 1, 1 + i, and i, the orientation of being in the counterclockwise direction, evaluate f z) dz. Solution 1 The positively oriented simple closed contour is given by the sum = 4 j=1 j, where 1 : z x 1 ) = x 1, x 1 1; : z y 1 ) = 1 + iy 1, y 1 1; 3 : z x ) = 1 x ) + i, x 1; 4 : z y ) = i 1 y ), y 1. 1

2 Thus, f z) dz = = + = 4 j= exp z) dz j exp z x 1 ) z x 1 ) dx 1 + exp z x ) z x ) dx + exp x 1 ) dx exp { [1 x ) i]} dx 1 1 exp exp i exp [ 1 iy 1 )] dy 1 1 z y 1 ) z y 1 ) dy 1 z y ) z y ) dy i exp [ i 1 y )] dy = exp x 1 ) x1=1 x exp [ 1 iy 1= 1)] y1=1 y 1= + exp { [1 x ) i]} x=1 x exp [ i 1 y = )] y=1 y = = [exp ) 1] {exp [ 1 i)] exp )} + {exp i) exp [ 1 i)]} [1 exp i)] = exp ) 1 exp ) exp i) + exp ) exp i) exp exp i) 1 + exp i) = exp ) = e ).

3 . Let R be the circle z = R R > 1), described in the counterclockwise direction. Show that Log z R z dz + ln R < R and prove Log z lim R R z dz =. Proof. We have R : z θ) = Re iθ, θ and Log z θ) z θ) = Log z θ) z θ) = < ln R + R, ln R) + θ where the latter equality follows from the fact that a + b < a + b if a, b >, R ln R) + ) R implying by the Theorem in Sec. 47 on the upper bounds for moduli of contour integrals the arclength of R is R) Log z R z dz ln R + R R Finally, since which implies = ln R +. R ln R lim R R = lim 1 R R = ln R + lim = R R then by the squeeze theorem for limits we conclude that Log z lim R R z dz =. This completes the proof. 3

4 3. Let be any simple closed contour which has the origin lying in its exterior. Let f be an entire function. Prove that 1 f dz =. z Proof. As f is an entire function and g z) = 1 z is analytic on the whole complex plane except for the isolated singularity at z = then the composition h z) = f g z)) is also analytic on the whole complex plane except for the isolated singularity at z =. Therefore, if be any simple closed contour which has the origin lying in its exterior then by the auchy-goursat Theorem in Sec. 5, we must have 1 = h z) dz = f dz. z 4

5 4. Let denote the positively oriented boundary of the square whose sides lie along the lines x = ± and y = ±. Evaluate the integral ) z z + 8) dz. Solution Let denote the positively oriented boundary of the square whose sides lie along the lines x = ± and y = ±. Then the function f z) = ) z + 8) is analytic on and in the interior of the positively oriented simple close contour which contains the origin in the interior which implies by the auchy Integral Formula see the Theorem in Sec. 54) that Therefore, 1 8 = f ) = 1 i f z) dz = z ) i z z dz = + 8) 8 = i 4. ) z z + 8) dz. 5

6 5. Let a function f be continuous on a closed bounded region R, and let it be analytic and not constant throughout the interior of R. Prove that f z) either has a zero in the interior of R or f z) has a minimum value in R which occurs only on the boundary of R. Proof. Let a function f be continuous on a closed bounded region R, and let it be analytic and not constant throughout the interior of R. Suppose that f z) for all z in the interior of R. We will now prove that f z) has a minimum value in R which occurs only on the boundary of R. Define the function g : R by g z) = 1 f z). Then g is continuous on a closed bounded region R and is not constant throughout the interior of R implying by the maximum modulus principle see the Theorem in Sec. 59) g z) = 1 fz) has a maximum in R which occurs only on the boundary of R. But this implies that f z) has a minimum value in R which occurs only on the boundary of R. Therefore, we have proven f z) either has a zero in the interior of R or f z) has a minimum value in R which occurs only on the boundary of R. 6

7 6. a) Obtain the Taylor series about z = 1 for the function f z) = ez z. b) Find the circle of convergence of this Taylor series. Solution 3 b) We know that f z) = ez z is analytic on the open disk {z : z 1 < 1}, but has a isolated singularity at z = which is on the boundary of the disk so that the f has a Taylor series at z = 1 and the circle of convergence of this series must be that disk for otherwise we could find a larger radius disk center at z = 1 for which the power series converges which would imply that z = 1 couldn t be a singularity of f, a contradiction). a) We know that the Maclaurin series for e z and 1 1 z are e z = 1 n! z < ), 1 1 z = z n z < 1). Thus, we can find from this the Taylor series about z = 1 for e z and 1 z to be e z = ee z 1 = a n z 1) n z 1 < ), where a n = e n!, 1 z = 1 1 [ z 1)] = where b n = 1) n. [ z 1)] n = b n z 1) n z 1 < 1), Hence, their product f z) = ez z has the Taylor series f z) = ez z = a n z 1) n b n z 1) n = c n z 1) n z 1 < 1), where the formula for the auchy product of these two series is given by See Sec. 73) e c n = a k b n k = k= k=k! 1)n k = e 1) n n 1 k! 1)k. n k= n 7

8 7. Prove that if f z) = { z ) when z ±, 1 when z = ±. Then f is an entire function. Proof. The function has no singularities in the whole complex plane z ) except possibly the isolated points z = ± where ) also has zeros. We also know that = sin z ), and implying that implying that for z ±, sin z = 1) n n + 1)! zn+1 z < ) = sin z ) = 1) n+1 z ) n+1 z < ) n + 1)! f z) = z 1 ) = z + 1 = 1) n+1 z + n + 1)! z z which the left-hand side is the product of two analytic function at z = with 1 1) n+1 z + n + 1)! ) n z ) n z= = 1 ) = f. This implies that f z) is analytic at z = as well. Similarly, we can show that f z) is analytic at z = since = sin z + ) = 1) n z + ) n+1 z < ) n + 1)! implying that for z ±, f z) = z 1 ) = z 1 = 1) n z n + 1)! z + z + ) n 8

9 which the left-hand side is the product of two analytic function at z = with 1 1) n z n + 1)! z + ) n z= = 1 = f ). This implies that f z) is analytic at z = as well. Therefore, we have proven f is analytic at each point in the complex plane and hence is an entire function. 9