How find $\max _{z: \ |z|=1} \ f \left( z \right)$ for $f \left( z \right) = |z^3 - z +2|$

Question

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

Answer 1

$\bf{My\; Solution::}$ 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.

Answer 2

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 $$|z^3 - z + 2|^2 =6-2\cos 2t+4\cos 3t-4 \cos t\Big|_{t = 2\pi/3}=6+1+4+2=13.$$