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math185f08-hw9sol

# math185f08-hw9sol - MATH 185 COMPLEX ANALYSIS FALL 2008/09...

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MATH 185: COMPLEX ANALYSIS FALL 2008/09 PROBLEM SET 9 For f : Ω C and n N , recall that g = f n is the function defined by g ( z ) = [ f ( z )] 2 for all z Ω. A function on Ω C is said to be meromorphic if it has only removable singularities or poles in Ω (i.e. no essential singularities or non-isolated singularities). 1. Let Ω C be a region and let f : Ω C . Suppose g = f 2 and h = f 3 are both analytic on Ω. Show that f is analytic on Ω. Solution. Note that g ( z ) = 0 if and only if h ( z ) = 0. So the zeros of g and h are in common. Let Z Ω be these common zeros. If Z contains a limit point, then g and h are identically zero by the identity theorem and f 0 is of course analytic. We will assume that all zeros in Z are isolated. Let z 0 Z . So there exist analytic functions g 1 and h 1 and some ε > 0 such that for all z D ( z 0 , ε ), g ( z ) = ( z - z 0 ) k g 1 ( z ) , g 1 ( z 0 ) 6 = 0 , h ( z ) = ( z - z 0 ) l h 1 ( z ) , h 1 ( z 0 ) 6 = 0 , where k and l are the orders of zero of g and h respectively at z 0 . But g 3 = f 6 = h 2 and so ( z - z 0 ) 3 k g 1 ( z ) 3 = ( z - z 0 ) 2 l h 1 ( z ) 2 . So we get g 1 ( z ) 3 h 1 ( z ) 2 = ( z - z 0 ) 3 k - 2 l . Since the lhs is non-zero for z = z 0 , the rhs is also non-zero for z = z 0 and this is only possible if 3 k - 2 l = 0. In other words, l > k , i.e. the order of zero of h is larger than the order of zero of g . So h/g has a removable singularity at z 0 . Since this is true for all z 0 Z , h/g may be extended to an analytic function e f on Ω. But since for all z Ω \ Z , e f ( z ) = h ( z ) g ( z ) = f ( z ) , and Ω \ Z clearly has limit points, e f f by the identity theorem. 2. Is there a polynomial p ( z ) such that p ( z ) e 1 /z is an entire function? Solution. By the Taylor expansion of exp, e 1 /z = X n =0 1 n !

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math185f08-hw9sol - MATH 185 COMPLEX ANALYSIS FALL 2008/09...

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