If function $f(z)$ is analytic and one valued in $|z-a|<R$ , prove that for $0<r<R$ $$f'(a)=\frac{1}{r\pi}\int_{0}^{2\pi}P(\theta)e^{-i\theta}d\theta$$
where $P(\theta)$ is real part of $f(a+re^{i\theta})$
EDIT : As pointed out by @Kavi Ram Murthy the question has been answered correctly here : Complex Analysis: Complex Integral Since the answer has not been accepted I will copy the answer and mark this as closed (with due credits to @Kavi Ram Murthy)
My approach : We know that $$f'(a)=\oint\frac{f(z)}{(z-a)^2}dz$$ subsituting $z=a+re^{i\theta}$ and simplifying i get
$$f'(a)=\frac{1}{2r\pi}\int_{0}^{2\pi}f(a+re^{i\theta})e^{-i\theta}d\theta$$
I am struck here I dont know how to procced further and prove first equation is equal to last equation . Any hints or sugestions ? (Also I may be be totally wrong in approaching the solution please let me know if I am
This answer was orginally answered by @Kavi Rama Murthy here Complex Analysis: Complex Integral
Use the fact that $\int f(a+re^{i\theta}) e^{i\theta } \, d\theta=0$ by Cauchy's Theorem. Take complex conjugate on both sides. Denote $\int g(a+re^{i\theta}) e^{-i\theta}\, d\theta$ by $I(g)$ . If u and v are the real and imaginary parts of f then Then $I(u−iv)=0$ Hence $I(u)=iI(v)$. Your answer is $\frac{1}{2πr}I(f)$ which is $\frac{1}{2πr}{I(u)+iI(v)}=\frac{1}{2πr}{I(u)+I(u))}=\frac{1}{πr}I(u)$. This is the given answer.