HW2sol - z and w in A the path from z to 1 and then from 1...

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P ROBLEM SET 2 Partial solution problems are 2. ( ) Let A be an open set. To show that A ∂A = it is enough to show that for any z A , z 6∈ ∂A . So let z A . Since A is open, z is an interior point of A . Hence there exists r > 0 such that B ( z,r ) A . But then B ( z,r ) ( C \ A ) = , which implies that z 6∈ ∂A . ( ) It is enough to show that if A is not open then A ∂A 6 = . So let us assume that A is not open. Then there exists a point z A which is not an interior point. That is for all r > 0 the disc B ( z,r ) is not a subset of A . This implies that for all r > 0 the intersection of B ( z,r ) with the complement of A is not empty. We also know that z A B ( z,r ) for any r > 0 , and hence A B ( z,r ) is not empty either. Thus z ∂A , and hence A ∂A 6 = . 3. An idea: Let A = B (0 , 3) \ B (0 , 2) = { z | 2 < | z | < 3 } and let E = { z | - 4 < Re z < 4 and - 1 < Im z < 1 } . 4. Let me just show how to prove that the set is connected. Note that for any z A = C \{ x | x R and x 0 } the line segment that connects z to 1 in the complex plane does not intersect the set of negative numbers, and is hence completely contained in A . Therefore for any two numbers
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Unformatted text preview: z and w in A the path from z to 1 and then from 1 to w is contained in A . Hence A is connected. 5. a) does not converge b) converges to ; Hint: use geometry to get that that z n = 2 tan-1 n c) converges to 6. We will prove the claim by contradiction. Assume that A is closed, that for all n , z n A , that lim n z n = w , but that w 6 A . Since the complement of A is an open set (by a theorem proven in class) w is an interior point of the complement. Hence there exists r &gt; such that B ( w,r ) A = . From the denition of a limit it follows that for = r there exists n such that for all n n it holds that | z n-w | &lt; r . Or in other words z n B ( w,r ) . Since z n A , this implies that B ( w,r ) A 6 = . Contradiction. 7. Example: Let A = B (0 , 1) , z n = 1-1 n and w = 1 ....
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