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Course: MATH MATH 552, Spring 2001
School: USC
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Analysis Spring Complex 2001 Homework V Solutions 1. Conway, chapter 4, section 5, problem 7. Let (t) = 1 + eit for 0 t 2. Find ( z )n dz for all positive integers n. z-1 2i By Corollary 5.8, this is (n-1)! times the n - 1rst derivative of f (z) = z n evaluated at z = 1. The n - 1rst derivative of z n is n!z and so the result is 2ni. 2. Conway, chapter 4, section 5, problem 9. Show that if f : C C is...

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Analysis Spring Complex 2001 Homework V Solutions 1. Conway, chapter 4, section 5, problem 7. Let (t) = 1 + eit for 0 t 2. Find ( z )n dz for all positive integers n. z-1 2i By Corollary 5.8, this is (n-1)! times the n - 1rst derivative of f (z) = z n evaluated at z = 1. The n - 1rst derivative of z n is n!z and so the result is 2ni. 2. Conway, chapter 4, section 5, problem 9. Show that if f : C C is continuous and analytic off the interval [-1, 1] then f is entire. This is an application of Morera's theorem. One has to show that the integral of f over the boundary of any rectangle is equal to zero. For rectangles which are disjoint from the interval, this follows from Cauchy's theorem. Rectangles which meet the interval can be broken into a union of subrectangles which lie on and above the x-axis and rectangles which lie on or below the axis. For any such rectangle which intersects the interval, show that the integral is given as the limit as 0 of the integrals over rectangles where the side lying on the horizontal axis is moved distance away from the axis. This is where continuity is used, since if f is continuous then f is uniformly continous, so f (x i) converges uniformly to f on compact intervals in x. Moreover, since f is bounded on compact sets, the integrals over the small vertical segments of length tend to 0, as they are bounded by times the maximum of |f (z)| over the rectangle. 3. Conway, chapter 4, section 6, problem 6. Let () = ei for 0 2 and () = 4 - for 2 4. Evaluate z2dz 2 . + Using partial fractions on the integrand z2 1 1 = 2 + 2i 1 1 - z - i z + i . Thus the integral is equal to 1 2i dz 1 - z - i 2i dz = n(; i) - n(; -i). z + i The point i lies in the unbounded component of the complement of so its index is 0. The curve is homotopic to a circle around -i traversed in counterclockwise orientation (write a seqeunce of elementary homotopies: push the bulge of the spiral in the first quadrant to a vertical segment on the imaginary axis; then push what is above the x-axis down to the x-axis; then what is left is star shaped from -i). Thus the index is 1, so the of value the integral is -1. 4. Conway, chapter 4, section 6, problem 10. Find all possible values of is any closed rectifiable curve not passing through i. dz 1+z 2 where 1 1 1 Using partial fractions, z21 = 2i z-i - z+i , so the integral breaks into a sum, each of +1 which is a constant times the index of with respect to one of the points i. Precisely, 1 the integral is equal to 2i 2i(n(; i) - n(; -i)). Thus the result is times an integer. Choosing to circles centered at i of small radius traversed k times in the positive or negative orientation, shows that any k for k = 0 can occur. Taking the circle of radius 4 centered at the origin, each index is equal to 1, so the difference is 0. Thus the set of possible values is all Z. 5. Show that an analytic branch of 1 - z 2 can be defined in any region G such that 1 and -1 lie in the same component of the complement, by proving that f dz is an f even multiple of 2i for f (z) = 1 - z 2 for every closed curve in G. -2z 1 1 1 1 Since f = 1-z2 = z+1 + z-1 , the integral f dz = z+1 + z-1 dz = 2i(n(; -1) + f f n(; 1)). It is given that 1 lie in the same component of the complement of G. They must then lie in the same component of the complement of , for if the complement of is U V with U and V open and disjoint and 1 lies in U while -1 lies in V , then Gc U V and so U Gc and V Gc separate 1 and -1 in Gc contrary to hypothesis. Thus the index of with respect to both points is the same, and f dz = 2n(; 1)2i f which is an even multiple of 2i. The argument to show that this permits a definition of the square root is then the same as for the example in class. 6. Show how the proof of Theorem 7.2 can be modified to evaluate f (z) h(z)dz f (z) where f, , and G are as in Theorem 7.2, and h(z) is analytic on G. Starting from the proof of Theorem 7.2, preserving notation, h(z) f h(z) h(z) h(z) g (z) = + + + + h(z) . f z - a1 z - a2 z - am g(z) h(z) 1 Then by Cauchy's integral formula, 2i z-aj dz = h(aj )n(; aj ) while since the integrand is analytic. Comnbining, the result is h(z) g dz = 0 g h(z) f (z) dz = f (z) m h(aj )n(; aj ). j=1
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