Exercise 11.  Let    be a nonconstant analytic function in the closed disk  .  

Suppose that    for  .  

Show that    has a zero in  .  

Hint:  Use both the maximum and minimum modulus principles.  

Solution 11.

See text and/or instructor's solution manual.

Solution.  (By contraposition)  If    does not have a zero, then    is analytic in  ,  

so by Theorem 6.15 (Maximum Modulus Principle) its maximum occurs on the boundary.  

Since    for    it follows that  

                       for   .

Which in turn implies that     for     .

The original hypothesis that    for    implies that    for  .

Thus we can infer that     for  ,   thus

                      for  .

Now apply Theorem 3.6 in Section 3.2 where we proved that if   for   then    is constant.  

Therefore we conclude that    is constant.

But this contradicts the statement that    be a nonconstant analytic function in the closed disk  .  

Since we have arrived at a contradiction, our statement  "  does not have a zero"  is false.

Therefore    must have a zero in  .  

This solution is complements of the authors.

(c) 2008 John H. Mathews, Russell W. Howell