<s>[INST] Find the maximum of the $| \left( w + 2 \right) ^3 \left( w - 3 \right)^2|$ with $|w|=1$ 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 
 [/INST] 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 $\phi = 0$
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