math185f08-hw6sol

math185f08-hw6sol - MATH 185: COMPLEX ANALYSIS FALL 2008/09...

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FALL 2008/09 PROBLEM SET 6 SOLUTIONS For α,β C , we will use the shorthand [ α,β ] to denote the line segment from α to β , parameterized by z : [0 , 1] C , z ( t ) = (1 - t ) α + . We will use the shorthand | z - a | = r to denote the circle of radius r centered at a , parameterized by z : [0 , 2 π ] C , z ( t ) = a + re it . 1. (a) Let a R . Show that lim r →∞ Z r - r e - ( x + ia ) 2 dx exists and is independent of a . Solution. First we will show that the limit exists. Observe that | e - ( x + ia ) 2 | = | e a 2 - x 2 +2 iax | = e a 2 e - x 2 . Since by Proposition 3.4 in the lectures ± ± ± ± Z r - r e - ( x + ia ) 2 dx ± ± ± ± Z r - r | e - ( x + ia ) 2 | dx = e a 2 Z r - r e - x 2 dx and the limit of the rhs exist, in fact Z -∞ e - x 2 dx = π (think probability distribution of N (0 , 1 2 )), we conclude that the limit lim r →∞ Z r - r e - ( x + ia ) 2 dx exists. Now consider the entire function f ( z ) = e - z 2 and choose a rectangular curve that the one we use in the proof of Cauchy’s theorem. Let Γ r = ∂R r = [ - r,r ] [ r,ia ] [ ia, - ia ] [ - ia, - r ], ie. R r is the rectangle with vertices - r + 0 i , r + 0 i , r + ia , - r - ia . Since f is analytic inside R r , Cauchy’s theorem implies that 0 = Z Γ r f = Z [ - r,r ] f + Z [ r,ia ] f + Z [ ia, - ia ] f + Z [ - ia, - r ] f = Z r - r e - x 2 dx + ie - r 2 Z a 0 e y 2 - 2 iry dy + Z r - r e - ( x + ia ) 2 dx - ie - r 2 Z a 0 e y 2 +2 iry dy. Note that as r → ∞ , then e - r 2 0. Hence, as r → ∞ , ± ± ± ± ± ie - r 2 Z a 0 e y 2 2 iry dy ± ± ± ± e - r 2 Z a 0 e y 2 dy 0 . Therefore, lim r →∞ Z r - r e - ( x + ia ) 2 dx = lim r →∞ Z r - r e - x 2 dx = π. Date
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math185f08-hw6sol - MATH 185: COMPLEX ANALYSIS FALL 2008/09...

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