Computing $\int_{\gamma} (z-a)^k dz$ without Cauchy's integral formula. $\gamma(t):=a+r\exp(it)$ with $r>0, a \in \mathbb{C}$

Question

how can I compute the following integral without using Cauchy's integral formula:

$k \in \mathbb{Z}$

$\gamma:[0, 2\pi] \rightarrow \mathbb{C},$

$\gamma(t):=a+r\exp(it)$ with $r>0, a \in \mathbb{C}$

$$\int_{\gamma} (z-a)^k dz$$

Answer 1

If $k\neq-1$, then, since $(z-a)^k$ has a primitive (namely, $\frac{(z-a)^{k+1}}{k+1}$) and $\gamma$ is a loop, the integral is $0$.

On the other hand$$\int_\gamma(z-a)^{-1}\,\mathrm dz=\int_0^{2\pi}\frac{ire^{it}}{re^{it}}\,\mathrm dt=2\pi i.$$

Answer 2

$\int_{\gamma}(z-a)^ndz=\int_0^{2 \pi}r^ne^{int}ire^{it}dt=r^{n+1}i\int_0^{2 \pi}e^{i(n+1)t}dt=\begin{cases}\ 2 \pi i & n=-1\\ 0 & n \neq -1\\ \end{cases}$