Hence lim z f z L 33 If z 6 1 3 i then the function h z z i z 13 i is

Hence lim z f z l 33 if z 6 1 3 i then the function h

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Hence lim z →∞ f ( z ) = L . 33. If z 6 = - 1 - 3 i then the function h ( z ) = z - i z +1+3 i is continuous in z . The discontinuity at z 0 = - 1 - 3 i is not removable because z 0 - i = - 1 - 3 i - i = - 1 - 4 i and z 0 + 1 + 3 i = 0. And we have lim z →- 1 - 3 i z - i z + 1 + 3 i = ( - 1 - 4 i ) lim z →- 1 - 3 i 1 z + 1 + 3 i = ( - 1 - 4 i ) = . 37. We use the definition of sin z using the exponential function. We have sin z = e iz - e - iz 2 i . From the Example 2.2.19 we know that e z is continuous everywhere in C . Since e iz = f ( g ( z )) for g ( z ) = iz and f ( z ) = e iz and both functions are continuous it follows from Theorem 2.2.13 that e iz is also continuous. Similar we show that the function e - iz is continuous. Then by Theorem 2.2.13 sin z = e iz - e - iz 2 i = 1 2 i e iz - 1 2 i e - iz is continuous in C . 41. (a) First, assume that f is continuous and A is open. Let z 0 be from f - 1 [ A ]. This means that f ( z 0 ) is in A . Since A is open there is ε > 0 such that the ε -neighborhood is a subset of A , or B ε ( f ( z 0 )) A . Since f is continuous at z 0 there is a δ > 0 such that if | z - z 0 | < δ then | f ( z ) - f ( z 0 ) | < ε . Thus f ( z ) is in A . And therefore z which is in f - 1 [ f ( z )] is in f - 1 [ A ]. Therefore the whole δ -neighborhood of z 0 is in f - 1 [ A ]. And it follows that f - 1 [ A ] is open. Second, assume that f - 1 [ A ] is open whenever A is open. Let z 0 be any complex number and ε > 0 be any too. Then by assumption the ε -neighborhood B ε ( f ( z 0 )) of f ( z 0 ) is open too. Therefore f - 1 [ B ε ( f ( z 0 ))] is open. In particular, there is δ > 0 such that B δ ( z 0 ) f - 1 [ B ε ( f ( z 0 ))]. It follows that for any z such that | z - z 0 | < δ we have | f ( z ) - f ( z 0 ) | < ε . Hence f is continuous. (b) First, assume that f - 1 [ A ] is closed whenever A is closed. To show that f is continuous by part(a) it is enough to show that f - 1 [ B ] is open whenever B is open. So, let open B be given. Since the complement of an open set is closed (problem 17 Section 2.1) we have A = C \ B is closed. Therefore, by assumption f - 1 [ A ] is closed too. Now we use the identity f - 1 [ B ] = C \ f - 1 [ A ] which is true for any set B and A = C \ B . Really, if z is in f - 1 [ B ]. Then f ( z ) is in B = C \ A . Therefore f ( z ) is in the complement of f - 1 [ A ]. On the other hand if z is in C \ f - 1 [ A ] then f ( z ) cannot be in A = C \ B . And, thus f ( z ) is in B . So, f - 1 ( z ) is in f - 1 [ B ]. Since f - 1 [ A ] is closed it follows from (2) and Problem 17 Section 2.1 that the complement set f - 1 [ B ] is open. Thus f is continuous. Second, assume that f is continuous and B is closed. Then A = C \ B is open. Therefore by part (a) it follows that f - 1 [ A ] is open. By (2) it follows that f - 1 [ B ] = C \ f - 1 [ A ]. f - 1 [ A ] is open. Then by Problem 17 Section 2.1 it follows that its complement, f - 1 [ B ] is closed.
56 Chapter 2 Analytic Functions Solutions to Exercises 2.3 1. If we take g ( z ) = 3 z 2 + 2 z and h ( z ) = z - 1 then we have f ( z ) = 3( z - 1) 2 + 2( z - 1) = g ( h ( z )) . By the chain rule and the formula that ( z n ) 0 = nz n - 1 for a positive integer n we have f 0 ( z ) = g 0 ( h ( z )) h 0 ( z ) = (6( z - 1) + 2)1 = 6( z - 1) + 2 = 6 z - 4 . 5. By the quotient rule and the formula for the derivative of a polynomial it follows 1 z 3 + 1 0 = (1) 0 ( z 3 + 1) - 1( z 3 + 1) 0 ( z 3 + 1) 2 = 0 - 3 z 2 ( z 3 + 1) 2 = - 3 z 2 ( z 3 + 1) 2 .

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