Evaluate $$\int_\gamma \bar{z}^2dz,$$ where $\gamma$ is the circle with centre $1$ and radius $1$ traced anticlockwise.
One parameterises the circle $\gamma$ as $z=1+e^{it}$ for $t\in[0,2\pi]$ and then solves the integral with $dz=ie^{it}dt$. So, $$\int_0^{2\pi}(1+e^{-it})^2(ie^{it})dt.$$
Why do we choose $z=1+e^{it}$? What is the general method to parameterise similar examples?
Thank you.
The key thing to know is that the map $[0, 2\pi] \to \mathbb{C}$, $t \mapsto e^{it}$ is an anticlockwise parameterisation of the unit circle, starting at $1$. We can shift and rescale this parameterisation to give an anticlockwise parameterisation of any circle. More precisely, the circle of radius $r$ centred at $a$ has an anticlockwise parameterisation given by $\gamma : [0, 2\pi] \to \mathbb{C}$, $\gamma(t) = a + re^{it}$. A clockwise parameterisation can be obtained by replacing $t$ with $-t$.