Contour integrals Evaluate $\int_C\dfrac{\mathrm{d}z}{z^2-1}$ where a) $C$ is the clockwise oriented circle $\left|z \right| = 2$; b) $C$ is the anti-clockwise oriented square with sides on $x= \pm2$ and $y= \pm2$; c) $C$ is the clockwise oriented circle $\left|z-1 \right|= 1$. So for this I would set $z = x+iy$ and separate the function into a real and imaginary part to solve the integral, right? And how do i know what bunds to put on the integral?

Note that $$\frac{1}{z^2-1}=\frac{1}{(z-1)(z+1)}$$ Hence, the function $f(z)=\frac{1}{z^2-1}$ has simple poles at $z=\pm 1$. We get $$\mathrm{res}_{z=1}f=\frac{1}{2},\quad\mathrm{res}_{z=-1}f=-\frac{1}{2}$$ For (a) the given contour encloses both poles so you have $$\int_{\gamma_a} \frac{1}{z^2-1}dz = -2\pi i\left(\frac{1}{2}+\frac{-1}{2}\right)=0$$ For (b) the contour is reversed, but also encloses both poles, so $$\int_{\gamma_b} \frac{1}{z^2-1}dz = 2\pi i\left(\frac{1}{2}+\frac{-1}{2}\right)=0$$ For (c) the contour encloses only the pole $z=1$, so $$\int_{\gamma_c} \frac{1}{z^2-1}dz = -2\pi i\cdot\frac{1}{2}=-\pi i$$