A question asks to solve the integral $\int_{\gamma} \frac{1}{z \bar{z}} dz$, where $\gamma \subset \mathbb{C}$ is a square centered at the origin with sides parallel to the axes.
Solution: Since the path is symmetric according to the origin, we could parameterize : $\gamma : [0,2] \to \mathbb{C}$ such that $\gamma(t+1)= - \gamma(t)$ for $t \in [0,1]$. Hence $$\int_{\gamma} \frac{1}{z \bar{z}} dz = \int_0^1 \frac{\gamma'(t)}{\gamma(t) \bar{\gamma(t)}}dt + \int_1^2 \frac{\gamma'(t)}{\gamma(t) \bar{\gamma(t)}}dt = \int_0^1 \frac{\gamma'(t)}{\gamma(t) \bar{\gamma(t)}}dt - \int_0^1 \frac{\gamma'(t)}{\gamma(t) \bar{\gamma(t)}}dt=0.$$
This is not clear what is the symmetry and why he defined $\gamma$ is that way. Does someone could explain to me those little subtle details? To know that the square is any.
Pick a vertex of the square, and choose a parameterisation $p : [0, 1] \to \mathbb{C}$ of the path to the opposite vertex along two of the sides. Then $\gamma : [0, 2] \to \mathbb{C}$,
$$\gamma(t) = \begin{cases} p(t) & 0 \leq t \leq 1\\ -p(t-1) & 1 \leq t \leq 2 \end{cases} $$
is a parameterisation of the square. Note that for $t \in [0, 1]$, $t + 1 \in [1, 2]$ so
$$\gamma(t+1) = - p((t+1) -1) = -p(t) = -\gamma(t).$$
By using a parameterisation with this symmetry, the integral could be split into two integrals which could be compared to one another. For an arbitrary parameterisation, you can't do this.