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 $\max _{z: \ |z|=1} \ f \left( z \right)$ ? 
 A: $\bf{My\; Solution::}$ Let $z=x+iy\;,$ Then $|x+iy| = 1\Rightarrow x^2+y^2 =1 $
Now We have To Maximize $$f(z) = \left|z^3-z+2\right| = \left|(x+iy)^3-(x+iy)+2\right|$$
We Get
$$f(x,y) = \left|x^3-iy^3+3ix^2y-3xy^2-x-iy+2\right|$$
$$f(x,y)=\left|x(x^2-3y^2)+iy(3x^2-y^2)-x-iy+2\right|\;,$$ Using $x^2+y^2 = 1$
$$f(x,y) = \left|x(4x^2-3)+iy(4x^2-1)-x-iy+2\right|=\sqrt{[(4x^3-4x)-2]^2+(4x^2y)^2}$$
Now Let 
$$g(x,y) = (4x^3-4x-2)^2+4x^4y^2 = (4x^3-4x-2)^2+4x^4(1-x^2)\;,$$ Using $x^2+y^2 = 1$
So Here we have to Maximize $$g(x) = (4x^3-4x-2)^2+4x^4-4x^6$$
Using Derivative Test, You Can Maximize It.
A: 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 $$|z^3 - z + 2|^2 =6-2\cos 2t+4\cos 3t-4 \cos t\Big|_{t = 2\pi/3}=6+1+4+2=13.$$
