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**Unformatted text preview: **is analytic everywhere except for singularities at the points z = 3 and z = 1, where its denominator vanishes. Since f ( z ) = h 1 ( z ) z 3 , where h 1 ( z ) = e z ( z + 1) 2 is analytic at z = 3 and h 1 (3) = 1 16 e 3 negationslash = 0, we conclude that z = 3 is a simple (order 1) pole. Similarly, f ( z ) = h 2 ( z ) ( z + 1) 2 , where h 2 ( z ) = e z z 3 is analytic at z = 1 with h 2 ( 1) = 1 4 e 1 negationslash = 0, we see that the point z = 1 is a double (order 2) pole. A complicated complex function can have a variety of singularities. For example, the function f ( z ) = 3 z + 2 e 1 / ( z 1) 2 z 2 + 1 (7 . 26) has simple poles at z = i , a branch point of degree 3 at z = 2, and an essential singularity at z = 1. As in the real case, and unlike Fourier series, convergent power series can always be repeatedly term-wise differentiated. Therefore, given the convergent series (7.22), we have the corresponding series f ( z ) = a 1 + 2 a 2 ( z z ) + 3 a 3 ( z z ) 2 + 4 a 4 ( z z ) 3 + = summationdisplay n = 0 ( n + 1) a n +1 ( z z ) n , f ( z ) = 2 a 2 + 6 a 3 ( z z ) + 12 a 4 ( z z ) 2 + 20 a 5 ( z z ) 3 + = summationdisplay n = 0 ( n + 1)( n + 2) a n +2 ( z z ) n , (7 . 27) and so on, for its derivatives. In Exercise you are asked to prove that the differentiated series all have the same radius of convergence as the original. As a consequence, we deduceseries all have the same radius of convergence as the original....

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