Please help me finding the value of the following integral. If $U\subseteq \mathbb{C}$ is an open set and $z_0\in U$ and $r>0$ and $\{z:|z-z_0|\le r\}\subseteq U$ and $j\in \mathbb{Z}^+$ $$\large \dfrac1{2\pi}\int_0^{2\pi}\left(z_0+re^{i\theta}\right)^jd\theta$$
I tried substituting $\left(z_0+re^{i\theta}\right) = z$ and I'm getting the upper and lower limits of the new integral same $\large z_{[lower]} = z_0 + r^{i0} = z_0+re^{i2\pi} = z_{[final]} $, thus making it zero. But the answer given in my book is $$ \large\dfrac1{2\pi}\int_0^{2\pi} u(z_0+re^{i\theta})d\theta = u(z_0)$$ where $u$ is a harmonic polynomial on $U$. So the answer becomes $$\large \dfrac1{2\pi}\int_0^{2\pi}\left(z_0+re^{i\theta}\right)^jd\theta = z_0^j$$
The result is easily proven with Cauchy's theorem. Set $z=e^{i \theta}$ and the integral becomes
$$\frac1{i 2 \pi} \oint_{|z|=1} \frac{dz}{z} (z_0+r z)^j $$
Expand the binomial out using the binomial theorem. Note that the only nonvanishing term in the integral corresponds to the first term by Cauchy's theorem. Thus the integral is equal to $z_0^j$ as was to be shown.