# Find the maximum value of $f(z)=|z^3-z+2|$ on the unit circle $|z|=1$

Let $$f:\mathbb{C} \rightarrow \mathbb{R}$$ be defined by $$f(z)=|z^3-z+2|\,.$$ What is the maximum value on the unit circle $$|z|=1$$ ?

My approach is as follows:

$$z=e^{i\theta}$$ as it is mentioned that the point lie on the unit circle.

$$f(z)=|1+\cos3\theta+i\sin3\theta +1-\cos\theta-i\sin\theta|\,.$$

Therefore, $$f(z)=|t|$$, where

$$t=2\cos^2\frac{3\theta}{2}+2i\sin\frac{3\theta}{2}\cos\frac{3\theta}{2}+ 2\sin^2 \frac{\theta}{2}-2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}\,.$$ $$t=2\cos\frac{3\theta}{2}\left(\cos\frac{3\theta}{2}+i\sin\frac{3\theta}{2}\right)-2i\sin \frac{\theta}{2}\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right)\,.$$

$$t=2\cos\frac{3\theta}{2}\,e^\frac{i3\theta}{2}- 2i\sin \frac{\theta}{2}\,e^\frac{i\theta}{2}\,.$$

I am not able to proceed from here.

Note: Please don't mark it as duplicate because the question in the website is same but my approach is different.

• Though this problem solution exist on this website, still would prefer to solve it from the step where I am not able to proceed. Sep 21, 2018 at 11:55
• Did you try Triple-angle Identities, i mean by $3*\frac{\theta}{2}$? Sep 21, 2018 at 12:23

Let $$z:=\exp(\text{i}\theta)$$ and $$t:=\cos(\theta)$$. Then, \begin{align}|z^3-z+2|^2&=\Big|\big(\cos(3\theta)-\cos(\theta)+2\big)^2+\text{i}\big(\sin(3\theta)-\sin(\theta)\big)\Big|^2 \\&=4\cos(3\theta)-2\cos(2\theta)-4\cos(\theta)+6 \\&=4(4t^3-3t)-2(2t^2-1)-4t+6 \\&=16t^3-4t^2-16t+8=:g(t)\,. \end{align} Thus, we want to maximize $$g(t)$$ subject to $$t\in[-1,+1]$$. Note that $$g'(t)=48t^2-8t-16=8(6t^2-t-2)=8(2t+1)(3t-2)\,.$$ That is, optimizing points of $$g(t)$$ in $$[-1,+1]$$ are $$t=-\dfrac12$$, $$t=\dfrac23$$, and the boundary points $$t\in\{-1,+1\}$$.
Note that $$g(-1)=4$$, $$g(+1)=4$$, $$g\left(-\dfrac12\right)=13$$, and $$g\left(\dfrac23\right)=\dfrac{8}{27}$$. We conclude that the minimum of $$g(t)$$ on $$[-1,+1]$$ is $$\dfrac{8}{27}$$, which is attained when $$t=\dfrac23$$, whereas the maximum of $$g(t)$$ on $$[-1,+1]$$ is $$13$$, which is attained when $$t=-\dfrac12$$. Translating this back to $$\theta$$, we see that $$\theta=\dfrac{2\pi}{3}$$ and $$\theta=\dfrac{4\pi}{3}$$ satisfy $$\cos(\theta)=-\dfrac12$$. Thus, $$z=\exp\left(\frac{2\pi\text{i}}{3}\right)=\dfrac{-1+\sqrt{3}\text{i}}{2}\text{ and }z=\exp\left(\frac{4\pi\text{i}}{3}\right)=\dfrac{-1-\sqrt{3}\text{i}}{2}$$ maximize $$|z^3-z+2|$$ for $$z\in\mathbb{C}$$ with $$|z|=1$$. The maximum value of $$|z^3-z+2|$$ for $$z\in\mathbb{C}$$ with $$|z|=1$$ is $$\sqrt{13}$$.
Similarly, the minimum value of $$|z^3-z+2|$$ for $$z\in\mathbb{C}$$ with $$|z|=1$$ is $$\sqrt{\dfrac{8}{27}}=\dfrac{2\sqrt{2}}{3\sqrt{3}}$$. The minimum is attained when $$z=\frac{2+\sqrt{5}\text{i}}{3}\text{ and }z=\frac{2-\sqrt{5}\text{i}}{3}\,.$$ (This happens when $$\theta=\text{arccos}\left(\dfrac23\right)$$ and when $$\theta=2\pi-\text{arccos}\left(\dfrac{2}{3}\right)$$.)