Evaluate $\int_{\lvert z\rvert = 2} \frac{\lvert dz\rvert}{\lvert z-1\rvert^2}$

Question

Let $z = 2 e^{i\theta}$, and I get $$\int_0^{2\pi} \frac 2{5-4cos\theta}{d\theta}$$ Then I have to use the integral-calculator to calculate its anti-derivative, and I get $$\frac {4arctan(3tan{\theta \over 2})}{3}$$ The final answer is $4\pi \over 3$, but I wonder if there exists a better way to calculate it.

Answer 1

This question has been tagged under "contour-integration". Following is one way to evaluate the integral as a contour one.

Parameterize the contour $|z| = 2$ by $z = 2e^{i\theta}$, we have

$$|dz| = 2d\theta = \frac{2dz}{iz}\quad\text{ and }\quad|z-1|^2 = (z-1)(\bar{z}-1) = (z-1)\left(\frac4z - 1\right)$$ The integral can be rewritten as

$$\int_{|z| = 2}\frac{|dz|}{|z-1|^2} = \frac{2}{i}\int_{|z|=2}\frac{dz}{(z-1)(4-z)}$$ Inside the contour, the integrand has only one simple pole at $z = 1$. This means $$\int_{|z| = 2}\frac{|dz|}{|z-1|^2} = \frac{2(2\pi i)}{i}{\rm Res}_{z=1}\frac{1}{(z-1)(4-z)} = \frac{4\pi}{4-1} = \frac{4\pi}{3} $$

Answer 2

use that $$\cos(x)=\frac{1-t^2}{1+t^2}$$ where $$t=\tan\left(\frac{1}{2}x\right)$$