1. ## Complex Variables

Let $f : D \rightarrow \mathbb C$ be holomorphic, let $z_0 \in D$ and let $r > 0$ be such that the disk $|w-z_0| \leq r \in D$ then $f'(z) = \frac{i}{\pi r} \oint\limits_0^{\pi} Im\left(f(z+re^{i\theta})e^{-i\theta}\right)d\theta$.
I started from the Cauchy integral formula and got $f'(z) = \frac{1}{2\pi r} \oint\limits_0^{2\pi} f(z+re^{i\theta})e^{-i\theta}d\theta$.
I feel like something is missing in the question.

2. Ok got it!

3. Originally Posted by vincisonfire
Ok got it!
Maybe you could post your solution so that others can enjoy it. You could post it inside spoiler tags.

4. After the exams. Promised.

5. We want to show $f'(z) = \frac{i}{\pi r} \oint\limits_0^{\pi} Im\left(f(z+re^{i\theta})e^{-i\theta}\right)d\theta$
Cauchy integral formula is $f'(z) = \frac{1}{2\pi i} \oint\limits_{\gamma} \frac{f(w)}{(w-z)^2}dw$
Let $w= z+r e^{i\theta}$ and $dw= ire^{i\theta}d\theta$.
$f'(z) = \frac{1}{2\pi i} \oint\limits_{\gamma} \frac{f(z+re^{i\theta})}{(z+re^{i\theta}-z)^2} ire^{i\theta}d\theta = \frac{1}{2\pi r} \oint\limits_0^{2\pi} f(z+re^{i\theta})e^{-i\theta}d\theta
$
.
Then, since f is holomorphic, $f'(z) = \sum\limits_{n=0}^{\infty} a_nz^n$ and from that we get that $\overline{f'(z)} = f'(\overline{z})$.
Now we integrate around the upper half of the circle from 0 to $\pi$ and around the lower half of the circle from 0 to $-\pi$. For the second we might as well integrate 0 to $\pi$ but change the sign of $\theta$.
$f'(z) = \frac{1}{2\pi r} \oint\limits_0^{\pi} f(z+re^{i\theta})e^{-i\theta}-\oint\limits_0^{\pi}f(z+re^{-i\theta})e^{i\theta}d\theta
$
.
$f'(z) = \frac{1}{\pi r} \oint\limits_0^{\pi} \frac{f(z+re^{i\theta})e^{-i\theta}-f(z+re^{-i\theta})e^{i\theta}}{2}d\theta
$

$f'(z) = \frac{1}{\pi r} \oint\limits_0^{\pi} iIm\left(f(z+re^{i\theta})e^{-i\theta}\right)d\theta
$

$f'(z) = \frac{i}{\pi r} \oint\limits_0^{\pi} Im\left(f(z+re^{i\theta})e^{-i\theta}\right)d\theta
$

I may have made mistakes though.