Let $C$ be the contour that goes once round the boundary of the square

$$\{z:-1 \leqslant \operatorname{Re} z \leqslant 1,-1 \leqslant \operatorname{Im} z \leqslant 1\}$$

in an anticlockwise direction. What is $\int_{C} \frac{d z}{z}$ ? Briefly justify your answer.

Explain why the integrals along each of the four edges of the square are equal.

Deduce that $\int_{-1}^{1} \frac{d t}{1+t^{2}}=\frac{\pi}{2}$.