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

Question

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

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

$\gamma(t):= \exp(it)$

$$\int_{\gamma} (z-a)^{-1} dz$$

Where $a \in \mathbb{C}$ & $0<|a|<1$.

It is clear from the winding number property that the integral should be $2πi$.

Please give me some hint.It will help me a lot.

Answer 1

We can evaluate the integral directly by straightforward analysis that reduces to an elementary integral from real analysis. We now proceed accordingly.

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Let $z=e^{i\phi}$ and $a=|a|e^{i\theta}$. Then, we can write

$$\begin{align} \oint_{|z|=1}\frac1{z-a}\,dz&=\int_{\theta}^{\theta+2\pi}\frac{ie^{i\phi}}{e^{i\phi}-a}\,d\phi\\\\ &=\int_{\theta}^{\theta+2\pi} \frac{ie^{i\phi}(e^{-i\phi}-a^*)}{1+|a|^2-2|a|\cos(\phi-\theta)}\,d\phi\\\\ &=\int_{\theta}^{\theta+2\pi} \frac{i-i|a|e^{i(\phi-\theta)})}{1+|a|^2-2|a|\cos(\phi-\theta)}\,d\phi\\\\ &=2i\int_{0}^{\pi} \frac{1-|a|\cos(\phi)}{1+|a|^2-2|a|\cos(\phi)}\,d\phi\tag1 \end{align}$$

Now evaluate the integral in $(1)$ by using, for example, the tangent half-angle substitution.

Can you finish now?