In proving $$f^{(n)}(w) = \frac{n!}{\rho^n \pi} \int_{0}^{2\pi} \Re ( f(w+\rho e^{i\theta})) e^{-in\theta} d\theta$$ where $f$ is analytic in $D(0;R)$, $0<r<R$, and let $w$ be an arbitary point in $D(0;r)$, I have some confusion.
The first idea is the Cauchy Integral Formula for derivatives, where we have $$f^{(n)}(w) = \frac{n!}{2\pi i} \int_{\partial D(w;\rho)} \frac{f(z)}{(z-w)^{n+1}} dz.$$
Plug in $ z = w + \rho e^{i\theta}$ we have $$f^{(n)}(w) = \frac{n!}{2\pi i} \int_{0}^{2\pi} \frac{f(w+\rho e^{i\theta})}{\rho^n e^{in\theta}} d\theta = \frac{n!}{2 \rho^n \pi i} \int_{0}^{2\pi} \frac{\Re(f(w+\rho e^{i\theta}))e^{i\theta}}{e^{in\theta}} d\theta,$$ and that's where I can go no further. I didn't know how to cancel the $2i$ in the denominator, or did the integrand has wrong power?
What is the trick?
Best Answer
With $ z = w + \rho e^{i\theta}$ we have $f^{n}(w) = \frac{n!}{2\pi i} \int_{0}^{2\pi} \frac{f(w+\rho e^{i\theta})}{\rho^{n+1} e^{i(n+1)\theta}} i\rho e^{i\theta} d\theta$ = $ \frac{n!}{2 \rho^n \pi } \int_{0}^{2\pi} {\{f(w+\rho e^{i\theta})\}e^{-in\theta}} d\theta$.