hw4sol - Complex Analysis Spring 2001 Homework IV Solutions...

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Complex Analysis Spring 2001 Homework IV Solutions 1. Conway, chapter 4, section 2, problem 10. Evaluate Z γ z 2 + 1 z ( z 2 + 4) dz where γ ( t ) = re it for t [0 , 2 π ] for all possible values of r , 0 < r < 2 and 2 < r < . Use partial fractions to write z 2 + 1 z ( z 2 + 4) = 1 4 1 z + 3 8 1 z - 2 i + 3 8 1 z + 2 i . Then Z γ z 2 + 1 z ( z 2 + 4) dz = 1 4 Z γ dz z + 3 8 Z γ dz z - 2 i + 3 8 Z γ dz z + 2 i . Now for γ the boundary of a disk D , oriented in the counterclockwise direction, we have proved that Z γ dz z - a = ± 2 πi, if a inside D 0 if a outside D. Applying this to the integrals above we have Z γ z 2 + 1 z ( z 2 + 4) dz = 2 πi 4 + 0 + 0 = πi 2 when 0 < r < 2 , and Z γ z 2 + 1 z ( z 2 + 4) dz = 2 πi 4 + 3 8 2 πi + 3 8 2 πi = 2 πi when 2 < r < . Before working on the next problem, be sure to review Prop. 1.6 in Chapter 3. 2. Conway, chapter 4, section 2, problem 13. This is a marathon problem – just one step after the other, but for a very long time. Since e z = n =0 z n n ! , e z - 1 = n =1 z n n ! so e z - 1 z = X n =1 z n - 1 n ! = X k =0 z k ( k + 1)! for z 6 = 0 . But the series on the right is convergent at z = 0 also, and so defines an analytic function. Since this formula shows that the series on the right is convergent for z with arbitrarily large modulus, the radius of convergence is
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hw4sol - Complex Analysis Spring 2001 Homework IV Solutions...

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