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461 MAT Assignment D - Solutions Posted November 12 (1) Assume f (z) is analytic on and inside the circle |z| = 4. Show that if f (z) = 0 for every z such that |z| = 4, then f (z) = 0 for every z such that |z| < 4 as well. (That is, show that if f (z) = 0 on the circle, then f (z) = 0 inside too.) By the Maximum Modulus Principle (or more specifically, the corollary on pg 171), the maximum value of |f (z)| in the circle |z| 4 must occur on the boundary |z| = 4. Since f (z) = 0 on the boundary, |f (z)| 0 inside the circle. This implies that f (z) = 0 inside the circle. Another way: Let C denote the circle and let z0 be a point interior to the circle. Then the conditions necessary for the Cauchy Integral Formula hold, and we can write f (z0 ) = 1 2i f (z) dz z - z0 C Since f (z) = 0 on C (and z - z0 = 0 on C), the integral on the right is zero and so f (z0 ) = 0. This works for all point inside C, so f (z) = 0 in the whole inside of C. (2) Find the Taylor series of the given function, expanded about the given point z0 . (a) f (z) = ez , z0 = i 2 ez = ez- 2 + 2 (z - i )n i 2 2 = e n! n=0 i i = n=0 i(z - i )n 2 n! (b) f (z) = z , 4 + z2 z0 = 0 z z 1 = +4 4 1 + z42 z = 4 = = n=0 z2 (-1) n=0 n z2 4 n z 4 n=0 (-1)n 2n z 4n (-1)n 2n+1 z 4n+1 1 NOTE: For questions (3),(4) and (5), C represents the circle radius 2, center 0, positively oriented. (3) (a) Find the Laurent series which represents f (z) = z sin in the region 0 < |z| < . sin( 1 ) = z2 = n=0 1 z2 (-1)n n=0 ( z12 )2n+1 (2n + 1)! 1 1 4n+2 (2n + 1)! z (-1)n 1 1 1 1 - + - + 2 6 10 z 3!z 5!z 7!z 14 1 1 1 (-1)n z sin( 2 ) = 4n+1 z (2n + 1)! z n=0 = = (b) Evaluate the following integral: z sin C 1 1 1 1 - + - + z 3!z 5 5!z 9 7!z 13 1 z2 Since dz the contour C is simple, closed, positive, around z = 0 and in the region where the Laurent Series from part 3(a) is valid, we can evaluate this integral by taking b1 1 (the coefficient of z ) and multiplying it by 2i. We see that b1 = 1. So z sin C 1 z2 1 dz = 2i (4) Evaluate the integral C ze z dz 1-z 1 by finding b2 (coefficient of 1 ) z2 ez valid for 1 < |z| < . in the Laurent expansion of 1-z e 1 z = n=0 1 1 n! z n 1 1 1 + + + 2 z 2!z 3!z 3 1 1 1 = 1-z z 1 -1 z = 1+ 2 = = 1 -1 1 z1- z 1 z n=0 -1 zn for 1 <1 z 1 1 1 - 2 - 3 - z z z 1 1 1 1 = - - 2 - 3 - 4 - z z z z 1 1 1 1 1 1 1 ez = (1 + + 2 + 3 + ) (- - 2 - 3 - ) 1-z z 2z 6z z z z -1 1 1 1 = + (-1 - 1) 2 + (-1 - 1 - ) 3 + z z 2 z 1 = z -1 - 1 By multiplying the series for e z and 1-z above, we see that b2 = -2 for the Laurent series valid for 1 < |z| < . Since C is simple, closed, positive, around |z| = 1 and in the region where our series is valid, we know that: 1 b2 1 z 1 = 2i C ez 1-z 1 (z - 0)1 dz C ze dz = 2i(-2) = -4i 1-z (5) Find the residue at z = 0 for the following function f (z) = Use this residue to evaluate the integral z 2 (3 1 - z) 1 z 2 (3 - z) C 1 1 1 = 3-z 31- 1 = 3 z 3 n=0 1 n z 3n z z2 z3 1 + 2 + 3 + 4 + = 3 3 3 3 1 1 1 1 z = + 2 + 3 + 4 + 2 (3 - z) 2 z 3z 3z 3 3 The above Laurent Series is valid for 0 < |z| < 3, so the residue at z = 0 is 1 . By 9 Cauchy's Residue Theorem, we know an integral around a closed simple positive contour 3 C is equal to 2i times the sum of the residues at the singularities inside C. z = 0 is the only singularity inside C (the other, at z = 3, is outside). So the integral z 2 (3 1 2i dz = - z) 9 C 4

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