Could you please explain to me how to find the maxima of $f(z)=\frac{1}{z^3+z^2+z+1}$ and the integral $\int_Cf(z)\mathrm dz$, where $C$ is an anticlockwise circle of radius $2$ centred around $(0,0)$?
For the maxima, I made the assumption they would be at the same points as for $z^3+z^2+z+1$ so I calculated $3z^2+2z+1=0$ giving $z=\frac{-2\pm\sqrt8}{6}$ and substituted both z into the original $f(z)$, is this correct, and is there a simpler way?
For the integral, I calculated $z^3+z^2+z+1=(z+1)(z+i)(z-i)$ but am not sure how to proceed from here.
We have a partial-fraction decomposition \begin{align}f(z)&=\frac{1}{2(z+1)}+\frac{1-z}{2(z^2+1)}\\ &=\frac{1}{2(z+1)}+\frac{i(1-z)}{4(z+i)}-\frac{i(1-z)}{4(z-i)} \end{align} We then obtain by cauchy's integral formula \begin{align} \oint_C f(z)dz=2\pi i\Big(\frac{1}{2}+\frac{i(1+i)}{4}-\frac{i(1-i)}{4}\Big)=0 \end{align}