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# hw8 - Mathematics 185 Intro to Complex Analysis Fall 2009 M...

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Mathematics 185 – Intro to Complex Analysis Fall 2009 – M. Christ Solutions to Selecta from Problem Set 8 3.1.2(a) Let c C and z 1 = c . Define z n +1 = z 2 n + c . Show that if | c | > 2 then lim n →∞ | z n | = . Comment. Our text says “lim n →∞ z n = ”. The most reasonable interpretation is the one I have given, replacing z n by | z n | , so that tending to infinity becomes well-defined. Solution. | z n +1 | = | z 2 n + c | ≥ | z n | 2 - | c | . Define r = | c | - 1. We will show by induction on n that | z n | ≥ | c | r n - 1 for all n 1. This holds for n = 1 since z 1 = c . If | z n | ≥ | c | r n - 1 then | z n +1 | ≥ | c | 2 ( r n - 1 ) 2 - | c | = | c | ( | c | r 2 n - 2 - 1) = | c | (( r + 1) r 2 n - 2 - 1) = | c | ( r 2 n - 1 + r 2 n - 2 - 1) ≥ | c | r 2 n - 1 ≥ | c | r n since r > 1 and 2 n - 1 n . This is the induction step. 3.1.4 Show that the series n =0 ( n 2 + z 2 ) - 1 converges at every point of C \ i Z , and that the convergence is uniform on every closed disk in this domain. Solution. The solution is motivated by the observation that ( n 2 + z 2 ) - 1 behaves like n - 2 for large n , so long as z is held constant. Note that ( n 2 + z 2 ) - 1 is defined if and only if n 2 + z 2 6 = 0. Given z , the n -th term of the series is defined for every n 0 if and only if z / i Z . | n 2 + z 2 | ≥ n 2 - | z | 2 by the triangle inequality. We cannot say that | ( n 2 + z 2 ) | - 1 n - 2 ; this is false if z is imaginary (unless | z | is sufficiently large). But we can hope to say that

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hw8 - Mathematics 185 Intro to Complex Analysis Fall 2009 M...

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