Parametrizing curve for complex analysis integral I'm trying to show that
$$
\int_{|z-z_0| = R} (z-z_0)^m \, dz = \begin{cases}0, & m \neq -1 \\ 2\pi i, & m =- 1. \end{cases}
$$
Here's my attempt at a solution:
We parametrize the curve at $z(\theta) = z_0 + Re^{i\theta}$ and therefore $dz = iRe^{i\theta} \, d\theta$. Substituting, we have
$$
\int_0^{2\pi} R^me^{im\theta} \, iRe^{i\theta} \, d\theta.
$$
However, I feel that this is wrong since there will be a dependence on $R$. Anyone have a suggestion?

For $m\geqslant 0, f(z) = (z - z_{0})^{m}$ is analytic on $\mathbb{C}$.  Thus the line integral is 0 ( by FTOC ).
For $m = -1, f(z) = 1$ is analytic on the closed disc $\rvert z - z_{0}\rvert\leqslant R$, so by Cauchy's integral formula this implies that $$1 = \frac{1}{2\pi i}\int_{\rvert z - z_{0}\rvert = R}\frac{1}{z - z_{0}}\,\mathrm{d}z.$$  Therefore the integral is $2\pi$.
For $m < -1$, take the derivative.. so $$0 = \frac{1}{2\pi i}\int_{\rvert z - z_{0}\rvert = R}(z - z_{0})^{m}\,\mathrm{d}z$$.
