How find $\max _{z: \ |z|=1} \ f \left( z \right)$ for $f \left( z \right) = |z^3 - z +2|$ Let $f : C \mapsto R $,  $f \left( z \right) = |z^3 - z +2|$. How find $\max _{z: \ |z|=1} \ f \left( z \right)$ ? 

i am going to parametrize the unit circle by $z = \cos t + i \sin t.$ we have $$\begin{align}|z^3 - z + 2|^2 &= (\cos  3t - \cos t + 2)^2 +(\sin 3t - \sin t)^2 \\&=\cos^2 3t + \cos^2t+4-2\cos 3t \cos t+4\cos 3t-4\cos t \\ &+\sin^2 3t + \sin^2 t-2\sin 3t \sin t\\
&=6-2\cos 2t+4\cos 3t-4 \cos t\end{align}$$
the critical points of $ |z^3 - z + 2|$ are given by
$$\begin{align}0 &=\sin 2t-3\sin 3t+\sin t \\
&= 2\sin t \cos t-3(3\sin t - 4 \sin^3 t) + \sin t\\
&=\sin t(2\cos t-9+12\sin^2 t+1)\\
&=-2\sin t(6\cos^2 t-\cos t-2)\\
&=-2\sin t(2\cos t+1)(3\cos t - 2) \end{align}$$
they are $$t = 0, 0.841, \pi, 2\pi/3, 4\pi/3,  5.442$$  by looking at these values, you can find the global max. it is at $t = 2\pi/3$ and the value is $$|z^3 - z + 2|^2 =6-2\cos 2t+4\cos 3t-4 \cos t\Big|_{t = 2\pi/3}=6+1+4+2=13.$$
