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Here's the question I'm trying to answer: Let $f$ be analytic on an open set $U$, let $z_{0}\in U$ and $f'(z_{0})\neq 0$. Show \begin{equation*} \frac{2\pi i}{f'(z_{0})}=\int _{C}\frac{1}{f(z)-f(z_{0})}dz \end{equation*} where C is a small circle centered at $z_{0}$.

I'm unsure how to start this problem. I've tried some manipulation of the Cauchy Integral formula, but haven't really gotten anywhere.

Additionally, this problem appears in a section where only one theorem is given, so I'm confident you need to use it but I can't figure out how. Here's the theorem:

Let $\{f_{n}\}$ be a sequence of analytic functions on an open set $U$, converging uniformly on every compact subset $K\subseteq U$ to a function $f$. Then $f$ is holomorphic. Furthermore, the sequence of derivatives $\{f'_{n}\}$ converges uniformly on every compact subset $K$ to $f'$.

Any hints would be appreciated. Thanks so much in advance.

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3 Answers 3

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Cauchy integral theorem says that if $f(z)$ is analytic inside a contour then

$\oint_\gamma \frac {f(z)}{z-z_0} \ dz = 2\pi i f(z_0)$

What can we do to get this integral into that form?

$\oint_\gamma \frac {z-z_0}{(z-z_0)(f(z) - f(z_0)} \ dz$

Let $g(z) = \frac {z - z_0}{f(z) - f(z_0)}$

$\oint_\gamma \dfrac {g(z)}{z-z_0} \ dz = 2\pi i g(z_0)$

So what is $g(z_0)$?

$f'(z_0) = \lim_\limits{z\to z_0} \frac {f(z) - f(z_0)}{z-z_0} = \lim_\limits{z\to z_0} \frac 1{g(z)}$

Since $f(z)$ is analyitc and $f'(z_0) \ne 0$ we can say $g(z_0) = \frac {1}{f'(z_0)}$

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  • $\begingroup$ Thanks so much, this makes a lot of sense. $\endgroup$
    – b17
    Mar 25, 2020 at 17:28
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Hint: $1/(f(z)-f(z_0))$ has a simple pole at $z_0$ and nowhere else inside the circle $C$ if $C$ is small enough. Use the residue theorem.

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Another approach would be to note that $f(z)-f(z_0) = f'(z_0)(z-z_0) + g(z) (z-z_0)^2$, where $g$ is analytic on an open set containing $U$ and for sufficiently small $|z-z_0|$ we have ${1 \over f(z)-f(z_0)} = {1 \over f'(z_0)(z-z_0)} \sum_{k=0}^\infty ( -{ g(z)(z-z_0) \over f'(z_0)} )^k$ and so $\int_C {dz \over f(z)-f(z_0)} = \int_C {1 \over f(z)'(z-z_0)} dz = 2 \pi i \operatorname{res}(\eta, z_0)$, where $\eta(z) = {1 \over f(z)'(z-z_0)}$.

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