Math 455            Solution to Selected Problems from Section 3.3
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1. (#10, page 123)
   First note that 
   
   Log(-z) + i \pi  = Log|z| + i Arg(-z) + i\pi, and

   log_0 (z) = Log|z| + i arg_0 (z).

   Hence, to identify Log(-z)+i\pi as log_0(z) one would have to check
   and identify the corresponding imaginary parts. So, for example let
   z be in the third quadrant. Then, -z will be in the first quadrant
   and Arg(-z) + \pi = 2\pi - Arg(z) will be in (0,2\pi]. Etc, you can
   check the remaining quadrants.
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2. (#13, page 123)
   Here all parts work the same, so I will do part (a). We want to find a
   branch of log(2z-1) that is analytic at all points in the plane except
   those on the ray {z| x \leq 1/2, y=0}.

   First, lets try the principal branch of the log, Log(2z-1).  This branch
   of the log is analytic at points in the plane except those that are 
   mapped into the ray {z| x \leq 0, y=0} by the function g(z)= 2z-1. Hence
   we need to identify this particluar set.

   2z-1 = (2x-1) + i(2y) implies

   2y = 0 => y=0
   2x-1 \leq 0 => x \leq 1/2.

   This calculation shows that Log(2z-1) is analytic all points in the complex
   plane except those on the ray {z| x \leq 1/2, y=0}. This is due to the
   fact that any point along this ray is mapped to the ray {z|x\leq 0, y=0}
   by function g(z)=2z-1. As you can see the principal branch of the log
   works in this case.
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3. (#14, page 123)
   In this problem we follow the given hint. Assume that there exists an
   analytic function F(z) in D such that F'(z) = 1/z for all z \in D. Now
   let z_o be a point in D\[-2,-1], i.e., z_o does not lie on the branch
   cut of Log z. Hence, G(z) = Log z is analytic at z_o with 

   G'(z_o) = d/dz(Log z) at z_o  = 1/z_o, 
   
   Therefore, we have showed that

   F' = G' for all z \in D\[-2,-1]. 

   Thus, F = G + c or F = Log z + c, where c is a constant. But here we get 
   a contradiction since Log z is not analytic on [-2,-1] and hence F.
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