380_hw8_solns[1]

# 380_hw8_solns[1] - this would be a contradiction Thus ± 1...

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Math 380 Hw 8 Solutions 1. Let G C : (a) G is a domain if and only if it is both open and connected. This is equivalent to G being both open and polygonally connected. (b) If G is a domain, it need not be simply connected. As an example, take G = C nf 0 g : Then G is open and connected. Let C denote the parametric curve given by z = e it with 0 ± t ± 2 Then C cannot be continuously deformed within G to a point. Thus G is not simply connected. (c) Here is an example of a simply connected open subset of the complex plane that is not connected: G := f z : j z ² 2 j < 1 g [ f z : j z + 2 j < 1 g : 2. Let D = C nf 0 g (a) We can parametriz ± 1 by z = e it , 0 ± t ± 2 and ± 2 by z = e it , 0 ± t ± 4 Now Z 1 1 z dz = Z 2 ± 0 e it ie it dt = 2 while Z 2 1 z dz = Z 4 ± 0 e it ie it dt = 4 Suppose ± 1 and ± 2 are continuous deformations of each other in D . Then the two integrals above would be equal by the Deformation Invariance theorem. Since the integrals have di/erent values,
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Unformatted text preview: this would be a contradiction. Thus ± 1 and ± 2 are not continuous deformations of each other in D: (b) No. Consider the example g ( z ) = 1 z 2 : In D , g ( z ) = f ( z ) with f ( z ) = & 1 z : Thus R j z j =1 g ( z ) dz = 0 : Note that lim z ! g ( z ) = 1 , so g cannot be analytic at z : 3. Let f : D ! C : (a) f is bounded on D if there exists a constant M such that j f ( z ) j ± M for all z 2 D: (b) f is an entire function if it analytic at every point of C : (c) No. An example of a bounded non-analytic function is f ( x + iy ) = 1 1+ x 2 : (d) The &aw is in the assertion lim z !1 e & z 2 = 0 : For example, lim y !1 e & ( yi ) 2 = lim y !1 e y 2 = 1 : 1...
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