# Uniform Limits of Analytic functions, complex analysis question

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.

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 $$g(z_0) = \frac {1}{f'(z_0)}$$

• Thanks so much, this makes a lot of sense.
– b17
Mar 25, 2020 at 17:28

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.

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 $$\eta(z) = {1 \over f(z)'(z-z_0)}$$.