Integration of $(z-a)^n$

Question

Let $C$ be the circle of radius $r$ centered around $a\in\mathbb{C}$, we're interested in finding the integral $$I_n = \int_C \frac{1}{(z-a)^n}\ \mathrm{d}z.$$ We know that $C$ can be parameterized by $C(t) = a+re^{it}$ where $t\in[0,2\pi]$. In the case that $n=1$, I can solve that $I_n=1$ as $$I_1 = \int_0^{2\pi}\frac{1}{(a+re^{it})-a}(ire^{it}) = 2\pi i.$$ However, for $n\ne1$, we have $$I_n = \frac{i}{r^{n-1}}\int_0^{2\pi}\frac{1}{(e^{it})^{n-1}}.$$ I know that is supposed to be zero, and I can prove it by using De Moivre's formula by expanding the exponential function. But I want to know if there's another way to proceed without invoking additional theorems? Thanks!

Answer 1

For $n\ne1$,

$$\int_0^{2\pi}\frac{ie^{it}}{(e^{it})^n}dt=i\int_0^{2\pi}e^{i(1-n)t}dt=\left.\frac{e^{i(1-n)t}}{1-n}\right|_0^{2\pi}=0.$$

The case of $n=1$ is trivial.

Answer 2

Hint: for $n \ne 1$, the function $\frac{1}{(z-a)^n}$ has an anti-derivative on $\mathbb C \setminus \{a\}.$