Find the maximum of the $| \left( w + 2 \right) ^3 \left( w - 3 \right)^2|$ with $|w|=1$

Question

Let $w \in \mathbb{C}$, and $\left | w \right | = 1$. Find the maximum of the function $| \left( w + 2 \right) ^3 \left( w - 3 \right)^2|$

Since $$|(w+2)^3(w-3)^2|=|w^5-15w^3-10w^2+60w+72|$$

Let $w=\cos x+i \sin x$. Then we have an ugly form

Answer 1

Hint:

$$\sqrt[5]{|w+2|^3\cdot|w-3|^2}\leq\dfrac{|w-3|+|w-3|+|w+2|+|w+2|+|w+2|}5$$

the equality occurs if $|w+2|=|w-3|$

Answer 2

Calling

$$ z_1=\rho_1 e^{i\phi_1} = \sqrt{(\cos\phi+2)^2+\sin^2\phi} = \sqrt{5+4\cos\phi}\\ z_2=\rho_2 e^{i\phi_2} = \sqrt{(\cos\phi-2)^2+\sin^2\phi} = \sqrt{10-6\cos\phi} $$

so

$$ |z_1^3z_2^2| = \rho_1^3\rho_2^2 = \left(5+4\cos\phi\right)^{\frac 32}|10-6\cos\phi| $$

now

$$ \frac{d}{d\phi}\left( \left(5+4\cos\phi\right)^{\frac 32}(10-6\cos\phi)\right) = 6 \sin\phi (4 \cos\phi+5)^{3/2}-6 \sin\phi (10-6 \cos\phi) \sqrt{4 \cos\phi+5}=0 $$

for $\phi \in \{0,\pm\frac{\pi}{3}\}$ so the maximum is

$$ 108=\max_{w}| \left( w + 2 \right) ^3 \left( w - 3 \right)^2|\;\;\mbox{s. t.}\;\;|w| = 1 $$

for $\phi = 0$