How to integral $\frac{1}{2\pi} \int_0^{2\pi} (z_0 + re^{i\theta})^j d\theta$

Question

Suppose $U\subseteq \mathbb{C}$ be an open set, Let $z_0 \in U$ and $r>0$ and assume $\{ z : |z-z_0|\leq r\}\subseteq U$, then how one can compute below without assuming cauchy integral formula? \begin{align} &\frac{1}{2\pi} \int_0^{2\pi} (z_0 + re^{i\theta})^j d\theta \\ &\frac{1}{2\pi} \int_0^{2\pi} \overline{(z_0 + re^{i\theta})}^j d\theta \end{align}

Answer 1

Hint: Binomial theorem. Then a simple calcultion plus some algebraic manipulation will give you the result.

Basically the "main result" you need is: $$\int_0^{2\pi} e^{i jt}dt=\frac{1}{ij}e^{ijt}\big|_0^{2\pi}=0,\quad j\neq 0$$ And $=2\pi$ if $j=0$.

Answer 2

Hint

If $C=\{z_0+re^{i\theta}\mid \theta\in [0,2\pi]\}$, your integrals are precisely $$-i\int_C \frac{z^j}{z-z_0}\mathrm d z\quad \text{and}\quad -i\int_C\frac{\overline{z}^j}{z-z_0}\mathrm d z.$$