$f(z)$ is analytical in $0<|z-a|<R$, and $r<R$ then prove that $$f'(a)=\frac{1}{\pi r}\int_{0}^{2\pi}P(\theta)\exp(-i\theta)d\theta$$ Where $P(\theta)$ is the imaginary part of $f(a+r\exp(i\theta))$
Using the cauchy integral formula $$f'(a)=\frac{1}{2\pi i }\int_{c}\frac{f(z)}{(z-a)^2}dz$$
We take $c$ as the circle of radius $r$ centered at $a$. this gives $z= a+r\exp(i\theta)$. We substitute above to get $$f'(a)=\frac{1}{2\pi r }\int_{0}^{2\pi}f(a+re^{i\theta})e^{-i\theta}d\theta$$
I am not able to proceed beyond this.