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Unformatted text preview: Math 185. Sample Answers to Problem Set #6 Page 153 1. In each case it suffices to show that the given function is analytic on the closed unit disc  z  1 . b . This function is entire, hence analytic on  z  1 . c . This function is analytic everywhere except for z satisfying z 2 +2 z +2 = 0 . By the quadratic formula, this equation is satisfied if and only if z = ( 2 4 8) / 2 = 1 i . But  1 i  = 2 , so 1 i lies outside of the closed unit disc; therefore 1 / ( z 2 +2 z +2) is analytic on the closed disc  z  1 . f . This function is analytic everywhere except for those z for which z + 2 is zero or a negative real number. Therefore it is analytic everywhere except for the real interval ( , 2) . This interval does not intersect the closed unit disc, so Log( z +2) is analytic everywhere on that disc. 2c. The function z/ (1 e z ) is analytic if and only if e z 6 = 1 , which holds if and only if z is not an integer multiple of 2 i . Thus it is analytic at z if and only if z / { 2 in : n Z } . Now if n = 0 then 2 in = 0 , and this is interior to C 2 , so it does not lie between the two contours. If n 6 = 0 , then  2 in  = 2  n  2 > 4 , so z is outside of C 1 . Therefore f is analytic everywhere between and on the two contours, so Corollary 2 in Section 46 applies, giving the desired equality of integrals. 4. a . The integral of e z 2 along the lower part of the given rectangle is Z a a e x 2 dx = 2 Z a e x 2 dx since e x 2 is an even function. Using the contour z ( x ) = x + ib , a x a , the integral along the upper part is Z a a e ( x 2 b 2 )+2 bxi ( 1) dx = e b 2 Z a a e x 2 cos 2 bxdx ie b 2 Z a a e x 2 sin2 bxdx = 2 e b 2 Z a e x 2 cos 2 bxdx since the first integrand is an even function and the second is odd.since the first integrand is an even function and the second is odd....
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 Fall '07
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