mt2-sample-ans

mt2-sample-ans - z = ± i with residues cos πi 2 i at z =...

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Math 185. Sample Answers to Second Sample Midterm 1. a . For a power series a n ( z - z 0 ) n , the circle of convergence is the largest circle with center z 0 such that the series converges for all z inside of the circle. b . Let f ( z ) be a function analytic in a punctured neighborhood 0 < | z - z 0 | < ± for some ± > 0 , and let n = -∞ c n ( z - z 0 ) n be the Laurent series for f in that neighborhood. Then the residue of f at z 0 is c - 1 . c . The Cauchy principal value of an integral R -∞ f ( x ) dx is the limit lim R →∞ Z R - R f ( x ) dx . 2. a . FALSE; for example if f ( z ) = 1 /z , if D = C \{ 0 } , and if C goes around the point z = 0 then the integral is nonzero. b . TRUE, by the maximum principle. c . TRUE, since having radius of convergence R is equivalent to a condition on the absolute values of the coefficients, and since taking real parts does not increase those absolute values. d . FALSE; for example f ( z ) = z and g ( z ) = 0 agree on the nonempty closed subset { 0 } of D = C , but they are not the same function. 3. The integrand has poles at
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Unformatted text preview: z = ± i , with residues cos πi/ 2 i at z = i and cos(-πi ) /-2 i at z =-i . Since these residues add up to zero (cosine is an even function) and both poles are inside the contour, the integral is zero. 4. Suppose that, for some z ∈ C and some ± > 0 , the open set | z-z | < ± is disjoint from the image of f . Let g ( z ) = 1 f ( z )-z . Then g is entire, and | g ( z ) | ≤ 1 /± for all z , so by Liouville’s theorem g is constant. This implies that f is constant, too. Thus we have shown the contrapositive of the desired result. 5. Using the series 1 1-z = ∞ X n =0 z n we have 1 z 2-1 =-1 1-z 2 =-∞ X n =0 z 2 n . Its radius of convergence is R = 1 , since the function 1 / ( z 2-1) is analytic in the disc | z | < 1 , but not at z = ± 1 . 1...
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This note was uploaded on 09/01/2010 for the course MATH 185 taught by Professor Lim during the Fall '07 term at Berkeley.

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