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?
 A: No, your work is correct. Factoring out constants, you simply need to evaluate
$$i R^{m + 1} \int_0^{2\pi} e^{i(m + 1) \theta} d\theta$$
Now if $m = -1$, then $R$ has exponent $0$ and will simply become $1$; if $m \ne -1$, then the integral vanishes and kills the dependence on $R$ anyways.
A: Use the cauchy integral formula
$\int_{|z-z_0| = R} \frac{f(z)}{(z-z_0)^{n+1}}dz=2\pi i\frac{f^{(n)}(z_0)}{n!}$
For $m=-1$, if we choose $f(z)=1$, then by cauchy integral formula  
$\int_{|z-z_0| = R} (z-z_0)^{-1}dz =2\pi i$
For $m\in\{-2,-3,-4,....\}$
Take $f(z)=1$
Then by cauchy integral formula we have
$\int_{|z-z_0| = R} (z-z_0)^{m}dz=2\pi i\frac{f^{(-m-1)}(z_0)}{(-m-1)!}=0\hspace{0.02cm} $$\hspace{0.02cm}$because derivatives of constant function are zero.
For $\hspace{0.05cm}m\in\{0,1,2,3,4,......\}$ $\hspace{0.02cm}$ you can use Cauchy Goursat Theorem to prove that
$\int_{|z-z_0| = R} (z-z_0)^{m}dz=0$
So in general 
$$
\int_{|z-z_0| = R} (z-z_0)^m \, dz = \begin{cases}0, & m \neq -1 \\ 2\pi i, & m = -1. \end{cases}
$$
A: For $m\geqslant 0, f(z) = (z - z_{0})^{m}$ is analytic on $\mathbb{C}$.  Thus the line integral is 0 ( by FTOC ).
For $m = -1, f(z) = 1$ is analytic on the closed disc $\rvert z - z_{0}\rvert\leqslant R$, so by Cauchy's integral formula this implies that $$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$$.
