MA424, S13 HW #6: Homework Problems 1. Answer the following, showing all work clearly and neatly. ONLY EXACT VALUES WILL BE ACCEPTED.

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1 MA424, S13 HW #6: Homework Problems 1 Answer the following, showing all work clearly and neatly. ONLY EXACT VALUES WILL BE ACCEPTED. NOTATION: Recall that C r (z) is the positively oriented circle with radius r and center z. Also, [z 0, z 1, z 2 ] is notation for the directed collection of line segments from z 0 to z 1 to z Find an antiderivative for each of the following functions and specify a domain on which the antiderivative is valid (you do not need to prove/justify your answer for the domain). (a) f(z) = ze z sin z (b) f(z) = 1 z 1 (c) f(z) = ze z2 1 z (d) f(z) = z 1/5 (principal value) (e) f(z) = Log z 2. Evaluate the given path integrals. Justify clearly the use of the FTC for Contour Integrals if you use it. You may cite and use calculations we did in class. (a) (b) (c) (d) [z 1,z 2,z 3 ] C 1 (0) (z 2 1) 2 z dz where z 1 = 0, z 2 = 1, z 3 = i ( ) (z 2 i) 2 i + z 2 i 3 (z 2 i) 2 sin 2 z dz where is any closed path. [z 1,z 2,...,z n] dz, where z 1, z 2,..., z n are arbitrary. dz

2 HW #6: Homework Problems MA424, S Complete the details of the proof of Cauchy s Theorem. From Calculus III, we have Green s Theorem: Theorem (GREEN S THEOREM). Let C be a positively oriented, piecewise-smooth, simple closed curve in the plane and let D be the region bounded by C. If P and Q have continuous partial derivatives on an open region that contains D, then ZZ I Q P P dx + Q dy = da. x y D C Now we have Cauchy s Theorem: Theorem (Cauchy s Theorem). Suppose that f is analytic and f 0 is continuous on and inside a simple closed contour. Then Z f (z) dz = 0. We will show that because of the work of Goursat that the hypothesis of f 0 being continuous is not needed and we get the Cauchy-Goursat Theorem in the text (page 151). Proof. (a) If we let f = u + iv, P = u and Q = v, (t) = x(t) + iy(t), dx = x0 (t) dt and dy = y 0 (t) dt, show Z i. Re f (z) dz Z Z ii. Im f (z) dz (u dx v dy), and = Z = (v dx + u dy).

3 MA424, S13 HW #6: Homework Problems 3 Also note that Then we have: f x = u x + i v x ( = Re and f y = u y + i v y ) ( ( + i Im ( = (u dx v dy) + i )) ) (v dx + u dy) (1) (2) (b) The hypotheses of Cauchy s Theorem gives us that the hypotheses of Green s Theorem are met. Use Green s Theorem to get double integrals equal to the integrals in (2). (c) Put your answers in #3b back into (2) and simplify to get one double integral of partial derivatives of f with respect to x and y. equal to (d) Rewrite the partials of f you get in #3c to be partials of u and v with respect to x and y. (e) Since f is analytic on and inside, the Cauchy-Riemann Equations are satisfied. Using your answer in #3d, show that from the equation in #3c that = = 0 and the theorem is proven.

4 MA424, S13 HW #6: Homework Problems 4 4. Complete the details of the proof Cauchy-Goursat Theorem (on triangular paths). Theorem (Cauchy-Goursat Theorem). (5- ) Let f be analytic ON AND INSIDE a simple closed contour. Then = 0. Γ We will prove this using the bisection method Goursat used ON A TRIANGULAR PATH without knowing anything about continuity of f to prove the theorem. It is more technical (and there are several methods) to prove it on a more general path, but these methods are much more understandable if you understand the triangluar path proof. Proof. Let be a triangular path with positive orientation around a triangle T and suppose f is analytic on and inside T. Divide T into 4 triangles T 1, T 2, T 3, and T 4 by connecting the midpoints of each of the sides. Orient the boundaries of these triangles positively and call these boundaries Γ 1, Γ 2, respectively. Let P be the perimeter of T and be the diameter (length of the longest side) of T. (a) Draw a picture of an example of T, T 1, T 2, T 3, and T 4 with the contours, Γ 1, Γ 2, Γ 3, and Γ 4 (and their directions) clearly labeled. With these orientations show/explain why 4 = and thus k=1 4 (b) There must be at least one T j (j = 1, 2, 3, or 4) for which 4 Γ j k=1 Rename this triangle T 1. Denote the perimeter of T 1 by P 1 and the diameter of T 1 by 1. Then i. P 1 = P and ii. 1 =.

5 MA424, S13 HW #6: Homework Problems 5 Divide T 1 as we did T into 4 smaller triangles to obtain T 2 with boundary Γ 2 : Γ 1 4 Γ 2 (c) Continuing on, one gets a sequence of triangles {T 1, T 2, T 3,... } with the following properties for any k N: T k+1 T k T, P k+1 = (a) P k, k+1 = (b) k, and 4 +1 Thus we have P k = (c) P k = (d) (e) Since each subsequent triangle is contained in the previous, and P k 0 as k, there exists z 0 T such that z 0 T k for all k (this is using the idea of a Cauchy sequence and completeness of the complex numbers from analysis, or argued in exercises #8 and #9 in 49). Let ε > 0. Since f is analytic on T, f differentiable at z 0, so by definitions of derivatives and limits, there exists δ > 0 such that if z z 0 < δ, f(z) f(z 0 ) f (z 0 ) < ε. (1) z z 0 (d) By doing algebra (show all work) on (1) to get rid of the fraction on the left hand side of the inequality, we get that if z z 0 < δ, < ε z z 0. (2)

6 MA424, S13 HW #6: Homework Problems 6 Since k 0, as k, there exists k N such that k < δ, and since z z 0 < k for any z T k, we have z z 0 < k < δ for any z T k. (e) If we let g(z) = 1 and h(z) = z, then i. g and h are ii. then by Cauchy s Theorem, we have for any k, g(z) dz = dz = (3a) iii. and h(z) dz = z dz =. (3b) iv. Thus we get (simplify integrand using (3a) and (3b), showing details) [ ] f(z) f(z 0 ) f (z 0 )(z z 0 ) dz =. (4) (f) Since z z 0 < k < δ, using (2) and the Theorem from 43, we get [ ] answer in (4) = f(z 0 ) f (z 0 )(z z 0 ) dz P k. (5) (g) Combining (5) with your answers from #3, we get (a) (b) P = (c). Since this is true for any ε > 0, we have = 0 and the theorem is proven.

7 MA424, S13 HW #6: Homework Problems 7 5. EXTRA CREDIT: WORTH UP TO 4 POINTS. Recall that log α z = ln z + i arg α z, where α < arg α z < α + 2π. Thus Log z = log π z. Consider the integral Log z dz, (t) = e it, 0 t π. Since Log z is not continuous at (π) = 1, it cannot be continuous in any domain containing the path and so we cannot apply the FTC for Contour Integrals directly. We must replace (carefully!) Log z with a different branch. (a) Show that Log z = log π z for all z on and explain why 2 Log z dz = log π z dz, 2 (b) Evaluate the integral using the FTC for Contour Integrals. (c) Come up with a different branch of log z that we could have used.