Proof using Cauchy's integral formula

  • #1
malawi_glenn
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Homework Statement


Let f and g ve analytic functions inside and on a simple connected contour [itex] \Gamma [/itex]. If f(z) = g(z) for all z on [itex] \Gamma [/itex], prove that
If f(z) = g(z) for all z inside [itex] \Gamma [/itex]


Homework Equations



[tex]
f(z_{0}) = \int_{\Gamma}\dfrac{f(z)}{z-z_{0}}dz [/tex]

if f is analytic in a simple connected domain containing [itex] \Gamma [/itex] and [itex] z_{0} [/itex] is a point insinde [itex] \Gamma [/itex].

The Attempt at a Solution



I know I must (?) prove that [itex]f(z_{0}) = g(z_{0}) [/itex] for all z_0, but i have no idea how to use the fact that f and g are equal on all point on [itex] \Gamma [/itex], I have no Lemma or Theoreme for this in my book (I am not allowed to use the theory of bounds for analytical functions, just Cauchy's integral formula)

Can someone give me a small hint ? =)
 

Answers and Replies

  • #2
NateTG
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Well, what happens if you write out:
[tex]f(z_0)-g(z_0)[/tex]
in the format you already have?
 
  • #3
malawi_glenn
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Well, what happens if you write out:
[tex]f(z_0)-g(z_0)[/tex]
in the format you already have?

okay i try that =) thanks!
 
  • #4
malawi_glenn
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[tex]
f(z_{0}) - g(z_{0}) = \dfrac{1}{2\pi i}\int_{\Gamma}\dfrac{f(z)}{z-z_{0}}dz - \dfrac{1}{2\pi i}\int_{\Gamma}\dfrac{g(z)}{z-z_{0}}dz[/tex]

I still dont know how to use the fact that they are equal for all z ON [tex]\Gamma[/tex] =(
 

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