hw6sol - Complex Analysis Spring 2001 Homework VI Due...

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Unformatted text preview: Complex Analysis Spring 2001 Homework VI Due Friday June 1 1. Conway, chapter 5, section 1, problem 1 b,h,i,j. Determine the nature of the isolated singularity at z = 0 of the following functions. If the function has a pole, find the singular part, for an essential singularity find the image of a small annulus. (b) f ( z ) = cos z z Using the series expansion for cos z , cos z = k =0 (- 1) k z 2 k (2 k )! we have f ( z ) = X k =0 (- 1) k (2 k )! z 2 k- 1 for z 6 = 0 . Since only one coefficient of a negative power of z is non-zero, f has a pole at z = 0 by corollary 1.18. The singular part is 1 z arising from the k = 0 term in the sum. (h) f ( z ) = 1 1- e z Since lim z (1- e z ) = 0, the limit as z 0 of | f ( z ) | = and so f ( z ) has a pole. From the series expansion for 1- e z =- k =1 z k k ! =- z k =1 z k- 1 k ! we see that 1- e z =- zb ( z ) where b ( z ) is analytic with b (0) = 1 . Thus zf ( z ) =- 1 b ( z ) has a removable singularity, and so f has a pole of order 1. Moreover, the singular part is- 1 z since the Taylor expansion of 1 b ( z ) starts with c = 1 b (0) . (i) The Taylor expansion of sin z converges for all z C, so sin 1 z = k =0 (- 1) k (2 k +1)! 1 z 2 k +1 . Then f ( z ) = X k =0 (- 1) k (2 k + 1)! 1 z 2 k which shows that infinitely many Laurent coefficients of negative index are non- zero, and thus by corollary 1.18 the function has an essential singularity at z = 0 . Application of the Casorati-Weierstrass theorem then says that the image of the annulus { z : 0 < | z | < } is dense. (j) This is similar to the preceding part. Again z n sin 1 z will have infinitely many non- zero Laurent coefficients of negative index, and so has an essential singularity. 2. Conway, chapter 5, section 1, problem 4. Find the Laurent expansion of f ( z ) = 1 z ( z- 1)( z- 2) in the three annuli ann (0; 0 , 1) , ann (0; 1 , 2) , ann (0; 2 , ) . This problem is most easily handled by partial fractions and manipulations with geo- metric series. A small simplification in this approach used by Abe Shepard is to note that the factor 1 z can be left, and partial fractions applied only to 1 ( z- 1)( z- 2) . Here 1 ( z- 1)( z- 2) = 1 1- z + 1 z- 2 and these two terms can be treated as follows. For | z | < 1 , 1 1- z has the geometric series expansion k =0 z k while 1 z- 2 =...
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hw6sol - Complex Analysis Spring 2001 Homework VI Due...

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