Compute $\int_{\gamma} \frac{1}{z-w}\mathrm dz, |w| \neq 1$ where $\gamma(t):=e^{it}, t \in [0, 2\pi]$

Question

I want to compute $$\int_{\gamma} \frac{1}{z-w}\mathrm dz,\quad\gamma(t):=e^{it}, t \in [0, 2\pi],\quad |w| \neq 1$$

What I tried so far:

$$\int_0^{2\pi} \frac{1}{e^{it}-w}ie^{it} \mathrm dt=\int_0^{2\pi} \frac{1}{1-\frac{w}{e^{it}}}i \mathrm dt = i \int_0^{2 \pi} \sum_{n=0}^\infty\frac{w^n}{e^{int}} \mathrm dt = i \sum_{n=0}^\infty \int_0^{2 \pi} \frac{w^n}{e^{int}} \mathrm dt = 2\pi i$$

if $\frac{|w|}{|e^{it}|}=|w|<1$.

Is that correct?

How is it done for$|w|>1$? Without Cauchy's integral formula

Thanks for your help!

Answer 1

For the $|w|>1$ case, you should proceed pretty much the way you started $$\int\limits_0^{2\pi} \frac{1}{e^{it}-w}ie^{it} \mathrm dt= -\frac{i}{w}\int\limits_0^{2\pi}\frac{e^{it}}{1-\frac{e^{it}}{w}}dt= -\frac{i}{w}\int\limits_0^{2\pi}e^{it}\left(\sum\limits_{n}\frac{e^{n\cdot it}}{w^n}\right)dt=\\ -\frac{i}{w}\int\limits_0^{2\pi}\left(\sum\limits_{n}\frac{e^{(n+1)\cdot it}}{w^n}\right)dt= -\frac{i}{w}\left(\sum\limits_{n}\int\limits_0^{2\pi}\frac{e^{(n+1)\cdot it}}{w^n}dt\right)=\\ -\frac{i}{w}\left(\sum\limits_{n}\frac{1}{w^n}\int\limits_0^{2\pi}e^{(n+1)\cdot it}dt\right)= -\frac{1}{w}\left(\sum\limits_{n}\frac{1}{w^n(n+1)}\int\limits_0^{2\pi}d\left(e^{(n+1)\cdot it}\right)\right)=\\ -\frac{1}{w}\left(\sum\limits_{n}\frac{1}{w^n(n+1)}\left(e^{(n+1)\cdot it}\biggr\rvert_{0}^{2\pi}\right)\right)= -\frac{1}{w}\left(\sum\limits_{n}\frac{1}{w^n(n+1)}\cdot \color{red}{0}\right)=0$$

Answer 2

Another way, using Cauchy's integral formula.

By Cauchy's integral formula, if $f(z)=1,$ for all $z\in \mathbb{C}$, then $$1=f(w) = \dfrac{1}{2\pi i} \int_{\gamma}\dfrac{f(z)}{z-w}dz = \dfrac{1}{2\pi i} \int_{\gamma}\dfrac{1}{z-w}dz, $$ as long $w$ is in the interior of the region defined by $\gamma.$

If $|w|>1$, then $\dfrac{1}{z-w}$ is holomorphic in the interior of the region defined by $\gamma$, then the integral is equal to 0.