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Sorry about the vague title, I'm not quite sure how to word it. Any edits would be very helpful!

Question:

Let $f(z)$ be analytic with $f'(z_0)\ne0$ where $z_0$ is a complex number Define $$g(w)= \frac{1}{2\pi i}\int_{\Gamma} \frac{z.f'(z)}{f(z)-w}\text{d}z$$

Calculate $g(f(z_0)$

My attempt:

So we would have $$g(f(z_0)= \frac{1}{2\pi i}\int_{\Gamma} \frac{z.f'(z)}{f(z)-f(z_0)}\text{d}z$$

Because we know $f'(z_0)\ne 0$ we can say that $f(z_0)$ is a simple pole? And therefore by Cauchy's Residue theorem we have

$$g(f(z_0)= Res(f,f(z_0))= f(z_0).f'(z_0)$$

Would this be correct?

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  • $\begingroup$ (wrote a mistake) by the residue theorem you would have, $g(f(z_0))= \frac{1}{2\pi i}\int_{\Gamma} \frac{z f'(z)}{f(z)-f(z_0)}\text{d}z = \sum_\rho Res( \frac{z.f'(z)}{f(z)-f(z_0)},\rho)$. where $\rho$ are the zeros of $f(z)-f(z_0)$ inside the contour $\Gamma$. if $z_0$ is inside the contour $\Gamma$ and $f(z)-f(z_0)$ has no other zeros, then $$g(f(z_0)) = Res(\frac{z f'(z)}{f(z)-f(z_0)}, z_0) = Res(\frac{z_0 f'(z_0)}{f'(z_0)(z-z_0)}, z_0) = z_0$$ $\endgroup$ – reuns May 24 '16 at 23:28
  • $\begingroup$ To me it doesn't look like $f(z_0)$ is a pole at all (the integrand doesn't have a singularity when $z=f(z_0)$. But it definitely looks like $z=z_0$ is a pole. What is the contour $\Gamma$, anyway? $\endgroup$ – Greg Martin May 24 '16 at 23:29
  • $\begingroup$ (maybe I shouldn't have written the whole solution) $\endgroup$ – reuns May 24 '16 at 23:31
  • $\begingroup$ $\Gamma \ \text{is a circle centred at} \ z_0$ $\endgroup$ – THISISIT453 May 24 '16 at 23:33
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    $\begingroup$ at $z = z_0$ clearly $\frac{1}{f(z) - f(z_0)}$ has a pole... for understanding it, you can also write the order 1 Taylor expansion $f(z)-f(z_0) \sim f'(z_0) (z-z_0)$ when $z \to z_0$, hence $\frac{1}{f(z) - f(z_0)} \sim \frac{1}{f'(z_0)(z-z_0)}$ $\endgroup$ – reuns May 24 '16 at 23:35

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