Transforming this integral

Question

Here is the integral. $0 < r < 1$

$\frac{1}{2\pi} \int_0^{2\pi} \frac{1-r^2}{1 + r^2 - 2r\cos{\varphi}} d\varphi$

I need to find a closed path In the complex plane I can rewrite this as.

Answer 1

The "integrating variable" is $\varphi$, so you can take the numerator $1-r^2$ out of it, and focus on rewriting $$\int^{2\pi}_{0}\frac{1}{1+r^2-2r\cos{\varphi}}d\varphi$$ over $\mathbb{C}$. As Dzoooks pointed out, the denominator is $(1-z)(1-\bar{z})$, however you want to eliminate the derivative of your curve, that comes along with the definition of complex integration. Dividing by $iz$ will do. So, taking $$f(z)=\frac{1}{(1-z)(1-\bar{z})iz}$$and $\gamma(\varphi)=re^{i\varphi}$, the integral on $\mathbb{C}$ equals$$\int_{\gamma}f(z)dz.$$Let's check it out:$$\int_{\gamma}f(z)dz=\int_{0}^{2\pi}\frac{1}{(1-re^{i\varphi})(1-re^{-i\varphi})ire^{i\varphi}}ire^{i\varphi}d\varphi=\int_{0}^{2\pi}\frac{1}{r^2-2r\cos{\varphi}+1}d\varphi$$ as we wished.