Math 455                      Homework                    Saff&Snider

Sec 4.6, p.167
__________________________________________________________________________

#11.  Assume that f(z_0)=0 for some z_0.  Consider the function
      g(z) = f(z) -1.  Then, g(z_0)=-1 and |g(z_0)| = 1.  But this
      is a contradiction since we are given that

      |g(z)| = |f(z) - 1| < 1  for all z in D.

__________________________________________________________________________

#17.  Let f(z) = (z-1)(z+1/2).  Then, f is analytic in disk of raduis 1
      centered at 0.  Hence, it takes its max mod on the boundary (by the
      max princ. thm).  So, we only need to maximize f on the boundary,
      where z= e^(it).

      On the boundary, values of f are:

      f(e(it)) = (e^it - 1)(e^it + 1/2)
	       = e^2it - 1/2 e^it - 1/2.

      Now, |f|^2 = (f) (f-conjugate).  Hence,

      |f|^2 = (e^2it - 1/2 e^it - 1/2) (e^(-2it) - 1/2 e^(-it) - 1/2)
	    = 3/2 - 1/2 cos t - cos 2t, where I used cos x=(e^ix-e^(-ix))/2
	    = 5/2 - 1/2 cos t - 2 cos^2 t, where I used cos 2x=2cos^2 x -1 

      Thus, we need to maximize the function -2 cos^2 t - 1/2 cos t + 5/2
      for t in [0, 2pi].  This is a mazimization problem from calculus.
      We have:

      g(t) = -2 cos^2 t - 1/2 cos t + 5/2

      g'(t) = 0  =>  t = 0, pi, 2pi, cos^(-1)(-1/8), 2pi-cos^(-1)(-1/8).

      Using the second derivative test we find that the max is at
      arccos(-1/8).

      Finally, plug in t = arccos(-1/8) into |f|^2, find the square of the
      max value, and then take the square root to find the max value of f.

____________________________________________________________________________