Suppose is analytic in some region containing B(0:1) and (2) = 1 where x1 = 1. Find a formula for 1. (Hint: First consider the case where f has no zeros in B(0; 1).) Exercise 7. Suppose is analytic in a region containing B(0; 1) and) = 1 when 121 = 1. Suppose that has a zero at z = (1 + 1) and a double zero at z = 1 Can (0) = ?

h(z) = g(z) for all z in the unit disk. In particular, h(0) = g(0) = -1, so 1(0) cannot be 1.By using the identity theorem for analytic functions,  

We know that if two analytic functions agree on a set that has a limit point in their domain, then they are identical.

Let g(z) = i/(z) - 1. Since i/(z)1 = 1 when |z| = 1, we can conclude that g(z) has a simple pole at z = 0 and no other poles inside the unit circle.

Suppose h(z) is analytic in the unit disk and agrees with g(z) at the zeros of i(z). Since i(z) has a zero of order 2 at z = 1, h(z) must have a pole of order 2 at z = 1. Also, i(z) has a zero of order 1 at z = i(1+i), so h(z) must have a simple zero at z = i(1+i).

Now we can apply the identity theorem for analytic functions. Since h(z) and g(z) agree on the set of zeros of i(z), which has a limit point in the unit disk, we can conclude that h(z) = g(z) for all z in the unit disk. In particular, h(0) = g(0) = -1, so 1(0) cannot be 1.

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