I need to calculate the maximum and minimum for the absolute value of the function
$$
f(z)=z^3-a
$$
for $z$ in the unit circle (center in origin) of $\mathbb{C}$ and $a\in\mathbb{C}, a\neq 0$. I tried to do this using some calculus argument but I ended up having to find the maximum and minimum of the function $\alpha \cos (3t)+\beta \sin (3t)$, where $t\in[0,2\pi]$ and $a=\alpha+i\beta$.
I'm now trying a geometric approach. Consider the line passing through $a$ and the origin in the plane. Since $g(z)=z^3$ restricted to the unit circle has its image on the unit circle, what I need to find are points $z$ in the domain of $g$ which images are the closer and farther from $a$ as possible. Obviosly those are the cubic roots of $\dfrac{-a}{|a|}$ and $\dfrac{a}{|a|}$, respectively.
Therefore, the maximum and minimum of $f$ over the unit circle would be $||a|+1|$ and $||a|-1|$, rescpectively.
Is this argument correct?
Best Answer
The argument is correct, but here is the (inferior) calculus version: write $a=re^{i\theta}$. Since $$|z^3-a|=|(e^{-i\theta/3}z)^3-r|$$ it follows that the maximum / minimum of $|z^3-a|$ are the same as for $|z^3-r|$. In terms of $z=e^{it}$, the square of $|z^3-r|$ is $$(\cos 3t-r)^2+\sin^2 3t = 1+r^2-2r\cos 3t \tag1$$ The extreme values of cosine are $\pm 1$, which correspond to $(1\pm r)^2$ in (1).