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Homework Help: Cauchy Integral Formula application

  1. Feb 19, 2012 #1
    [itex]f[/itex] is analytic on an open set [itex]U[/itex], [itex]z_0\in U[/itex], and [itex]f'(z_0)\neq 0[/itex]. Show that

    [tex]\frac{2\pi i}{f'(z_0)}=\int_C\frac{1}{f(z)-f(z_0)}dz[/tex]

    where $C$ is some circle center at $z_0$.

    S0 ,[itex]f(z)-f(z_0) = a_1(z-z_0)+a_2(z-z_0)^2+\cdots[/itex] with [itex]a_1=f'(z_0)\neq 0[/itex]. But why can [itex]f(z)-f(z_0)[/itex] be expanded this way?
     
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  3. Feb 19, 2012 #2

    Dick

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    You are given that f(z) is analytic. Doesn't that mean it can be locally expanded in a power series?
     
  4. Feb 19, 2012 #3
    Yes but I don't understand how the expansion of [itex]f(z)-f(z_0)[/itex] is that expansion.
     
  5. Feb 19, 2012 #4

    Dick

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    f(z) is analytic at z=z0 if f(z)=a0+a1*(z-z0)+a2*(z-z0)^2+... in a neighborhood of z0? What's f(z0)?
     
  6. Feb 19, 2012 #5
    The accumulation point.
     
  7. Feb 19, 2012 #6

    Dick

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    No, f(z0)=a0.
     
  8. Feb 27, 2012 #7

    The Taylor Series expansion of [itex]f(z) = \sum_{n = 0}^{\infty}c_n(z - z_0)^n[/itex] [itex]= c_0 + c_1(z - z_0) + c_2(z - z_0)^2 +\cdots[/itex], and [itex]f(z_0) = c_0[/itex].
    So,
    $$
    f(z) - f(z_0) = c_1(z - z_0) + c_2(z - z_0)^2 +\cdots.
    $$
    By factoring, we obtain [itex]f(z) - f(z_0) = c_1(z - z_0)\left[1 + \frac{c_2}{c_1}(z - z_0)^2 +\cdots\right][/itex].
    Then
    $$
    \frac{z - z_0}{f(z) - f(z_0)} = \frac{1}{c_1 + c_2(z - z_0) + \cdots}.
    $$
    So [itex]g(z) = \frac{z - z_0}{f(z) - f(z_0)}[/itex] is analytic on a disc (what is the justification for this part-analytic?) at [itex]z_0[/itex] (How does this disc relate to C?) and where the value of [itex]z_0[/itex] is redefined as [itex]1/c_1[/itex].
    From Cauchy's Theorem, we have
    $$
    \frac{1}{2\pi i}\int_C\frac{g(z)}{z - z_0} = g(z_0) = \frac{1}{2\pi i}\int_C\frac{dz}{f(z) - f(z_0)} = \frac{1}{f'(z_0)}
    $$
    hence
    $$
    \int_C\frac{dz}{f(z) - f(z_0)} = \frac{2\pi i}{f'(z_0)}.
    $$
     
  9. Feb 27, 2012 #8

    Dick

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    Your g(z) has a removable singularity at z=z0. And the problem just says "some circle centered at z0". Take the circle small enough so that g(z) is analytic in the circle.
     
  10. Feb 27, 2012 #9
    Ok thanks.
     
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