Cauchy Integral Formula application

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[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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  • #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?
 
  • #3
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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?
Yes but I don't understand how the expansion of [itex]f(z)-f(z_0)[/itex] is that expansion.
 
  • #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)?
 
  • #5
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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)?
The accumulation point.
 
  • #6
Dick
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  • #7
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No, f(z0)=a0.

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
$$
\int_C\frac{dz}{f(z) - f(z_0)} = \frac{2\pi i}{f'(z_0)}.
$$
 
  • #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.
 
  • #9
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Ok thanks.
 

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