I wish to show:
$$\left| \int_{\Gamma} \frac{1}{z^2-1} dz\right|~\leq \frac{3\pi}{16}$$ where $\Gamma$ is the arc of the circle given by $|z|=3$ contained in the first quadrant.
Here is what I have got so far:
$$\left| \int_{\Gamma} \frac{1}{z^2-1} dz\right|~\leq ML$$ by the ML lemma. Now it is clear that the length of $\Gamma$ is going to be $\frac{3\pi}{2}$ just by using the circumference formula for circles and dividing by $4$ so now I have that
$$\left| \int_{\Gamma} \frac{1}{z^2-1} dz\right|~\leq \frac{3\pi M}{2}$$
but I am struggling to find the maximum value that $f(z)=\frac{1}{z^2-1}$ takes over $\Gamma$.
Any hints?
At least when performing estimates in residue theorem calculations, you don't need to actually find the maximum, you only need an upper bound. Here there is an easy one to come by:
$$\frac{1}{|z^2-1|} \leq \frac{1}{|z^2|-1}=\frac{1}{9-1}=\frac{1}{8}.$$
Here all I've used is the triangle inequality. (As it turns out this is the maximum, but you don't need to prove that for your problem.)