Complex Analysis: Complex Integral

Question

If the function $f(z)$ is analytic and one valued in $|z-a|<R$, How do we prove that $0<r<R$, $f^{'}(a)$ = $\frac{1}{{\pi}r}\int P(\theta)e^{-i\theta}d\theta$, where $P(\theta)$ is the real part of $f(a+re^{i\theta})$. I am guessing Cauchy's integral formula can be used but that has lead me to $f^{'}(a)$ = $\frac{1}{2{\pi}r}\int f(a+ re^{i\theta})e^{-i\theta}d\theta$. How to proceed from here.

Answer 1

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 $\frac 1 {2\pi r} I(f)$ which is $\frac 1 {2\pi r} \{I(u)+iI(v)\}=\frac 1 {2\pi r} \{I(u)+I(u))\}=\frac 1 {\pi r} I(u)$. This is the given answer.