Evaluate $\int_{0}^{2\pi}f(z_0+re^{i\theta })e^{ki\theta }d\theta $

Question

Let $f$ be entire evaluate $\int_{0}^{2\pi}f(z_0+re^{i\theta })e^{ki\theta }d\theta $ , $ k \in \mathbb{N}$ $k\geq1$

I tried using Cauchy theorem because the integral looks like a line integral on a closed curve, but first I think I need to show that $(1)e^{kiθ}=e^{iθ}$ when $k \in \mathbb{N}$ $k\geq1$

$(1)\Rightarrow cos(kθ)=cos(θ) $ and $sin(kθ)=sin(θ)\Rightarrow$ $kθ-θ=2n\pi, n\in \mathbb{Z}$ (this is where I stuck, i know this is easy but how can i show $e^{kiθ}=e^{iθ}$?)

after proving (1) we have the integral is equal to $\int_{0}^{2\pi}f(z_0+re^{i\theta })e^{i\theta }d\theta $ $=\frac{1}{ir}\int_{γ}f(z)dz$ where $γ(θ)=z_0+re^{i\theta }$ , $θ \in [0,2\pi]$ then all the requirements for the Cauchy integral are true so its equal to $0$

Answer 1

Your calculation is not correct, $e^{ki \theta}=e^{i\theta}$ does not hold.

But you can apply Cauchy's integral theorem to $$ \int_\gamma f(z)(z-z_0)^{k-1} \, dz = \int_0^{2 \pi} f(z_0+re^{ki \theta)})r^{k-1}e^{(k-1)i \theta} ir e^{i \theta} \, d\theta = ir^k \int_0^{2 \pi} f(z_0+re^{ki \theta})e^{ki \theta} \, d\theta $$

Answer 2

Why not use Integration by parts? $\int uv \ dx = u \int vdx - \int \left(\frac{du}{dx} \int vdx \right) dx$. For $u = f(z_0 + re^{i\theta}), v = e^{ki\theta}, x = \theta$, we can expand the integral, $$\begin{aligned} \int_{0}^{2\pi} f(z_0 + re^{i\theta})e^{ki\theta}d\theta &= f(z_0 + re^{i\theta}) \int_{0}^{2\pi} e^{ki\theta}d\theta - \int_{0}^{2\pi} \left( \frac{d f(z_0 + re^{i\theta})}{d\theta} \int_{0}^{2\pi} e^{ki\theta}d\theta \right) d\theta \\ \int_{0}^{2\pi} e^{ki\theta}d\theta &= \frac{1}{ki} (e^{ki2\pi} - e^{ki0}) = 0 \\ \therefore \int_{0}^{2\pi} f(z_0 + re^{i\theta})e^{ki\theta}d\theta &= f(z_0 + re^{i\theta}). 0 - \int_{0}^{2\pi} \left( \frac{d f(z_0 + re^{i\theta})}{d\theta} .0 \right) d\theta = 0 \end{aligned}$$