Math 455                          Homework                 Saff&Snider

Sec 4.5, p.160
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#10.  Just apply the CIF to both sides.

      Note that the right hand side is 2ipi f'(z_0), by applying the
      CIF for derivatives.

      Now, note that the left hand sidei is also 2ipi f'(z_0), by 
      applying the CIF to the function f'(z).       

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#15.  This problem is an application of Thm 15 in this section.  First
      define a new function G by

      G(z) :=  (1/2pi i) int_{gamma} g(w)/(w-z) dw,

      where gamma = {w: |w| = 1} and g(z) := f(z)/z.

      Note that g is continuous on gamma.  Hence, by Thm 15, we have
      that G(z) is analytic at every point z NOT on gamma.

      Claim:  G(z) = F(z) for all z: |z| < 1.

      Pf.  The hint in your beek is really not necessary since we can
      directly apply the CIF to G.  But below is the soln using the hint
      in your text.  Note that

      g(w) / (w-z) = f(w) / w(w-z) =

		   =  (1/z) [f(w)/(w-z) - f(w)/w], partial fraction

      Hence,

      G(z) = (1/z) [(1/2ipi) int_{gamma} f(w)/(w-z) - (1/2ipi) int_{gamma}

	     f(w) / w

	     = (1/z) [f(z) - f(0)], by the CIF

	     = f(z)/z

	     = F(z), where I used the fact that f(0)=0.

      Therefore, G(z) = F(z) for all z \neq 0.  Furtheremore, 

      G(0) = (1/2ipi) int_{gamma} f(w)/w^2 = f'(0) = F(0), by CIF.

      Hence, F(z) = G(z) for all z in this disk, and since G is analytic,
      so is F there.
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