Here is a picture of the problem (sorry I have trouble typing the symbols). Suppose f is holomorphic inside of a simple closed curve gamma. Show that $$f'(a)=\frac{1}{2\pi}\int_0^{2\pi}\mathrm{e}^{-\mathrm{i}\theta}f(a+\mathrm{e}^{\mathrm{i}\theta})\mathrm{d}\theta$$
I started by letting $z=a+\mathrm{e}^{\mathrm{i}\theta}$ and $\frac{\mathrm d z}{d\theta}=\mathrm i\mathrm{e}^{\mathrm{i}\theta}=\mathrm iz$ so $\mathrm d\theta=\frac{\mathrm dz}{\mathrm iz}$. Then we have $$ \frac{1}{2\pi}\int \frac{f(z)}{z-a}\frac{\mathrm dz}{\mathrm iz}= \frac{1}{2\pi\mathrm i}\int \frac{f(z)}{z(z-a)}\mathrm dz $$ I'm not entirely sure where to go from here. I looks like we should use the Cauchy integral formula, but I'm not sure how with the extra $z$ in the denominator.
Thank you!
Use the general version of Cauchy Integral Formula:
$$f^{(n)}(a) = \frac{1}{2 \pi i} \int_C \frac{f(z)}{(z-a)^{n+1}} dz$$
For the first derivative you get exactly $(z-a)^2$ in the denominator.
By the way The problem is not true as stated. Since you need to integrate over a circle of radius 1, you need your function to be Analytic on a closed disk of radius at least 1, if your curve is too close to a, and the function is not analytic outside $a$ the problem is not true. Note that as given, the problem doesn't guarantee that $f(a+e^{it})$ even makes sense.