<s>[INST] Prove that if $ z$ is a complex number such that $ |z| \leq 1 $ then $|z^2 -1|\cdot |z-1|^2 \leq 3 \sqrt{3}$. Prove that if $ z$ is a complex number such that $ |z| \leq 1 $ then $|z^2 -1|\cdot |z-1|^2 \leq 3 \sqrt{3}$.
I tried geometric solution or using triangle inequality but doesn't work  because I lose  equality case.
 [/INST] We have $|z^2 -1|\cdot |z-1|^2 = \lvert p(z) \rvert$ with $p(z) = (z^2-1)(z-1)^2$. Since $p(z)$ is holomorphic, $ \lvert p(z) \rvert$ attains its maximum on the closed disk $D^2 = \{ z \in \mathbb C \mid \lvert z \rvert \le 1\}$ on the boundary $\partial D^2 = \{ z \in \mathbb C \mid \lvert z \rvert = 1\}$. This follows from the maximum modulus principle.
Each $z$ with $\lvert z \rvert = 1$ has the form
$$z = x \pm i \sqrt{1-x^2} .$$
We get
$$z^2 -1 = x^2 -(1-x^2) \pm 2i x  \sqrt{1-x^2} -1 = 2(x^2 -1 \pm ix \sqrt{1-x^2}), $$
$$\lvert z^2 -1 \rvert = 2\sqrt{(x^2-1)^2 + x^2(1-x^2)} = 2\sqrt{1-x^2},$$
$$z -1 = x -1 \pm i \sqrt{1-x^2} ,$$
$$|z-1|^2 = (x-1)^2 + 1- x^2 = 2(1-x) .$$
Thus
$$\lvert p(z) \rvert  = 4\sqrt{1-x^2}(1-x) .$$
It therefore suffices to find the maximum of $q(x) = \sqrt{1-x^2}(1-x)$ for $x \in[-1,1]$. Since $q(x)$ is differentiable in $(-1,1)$, the maximum is either attained at one of the boundary points $x = \pm 1$ of the interval $[-1,1]$ or at a point $x \in (-1,1)$ with $q'(x) = 0$. Since $q(-1)= q(1) = 0$ and $q(0) = 1$, we have to find $x$ with $q'(x) = 0$. We get
$$q'(x) = \frac 1 2 (1-x^2)^{-1/2}(-2x)(1-x) + \sqrt{1-x^2}(-1) = \frac{2x^2 -x -1}{\sqrt{1-x^2}} .$$
We have $2x^2 -x -1 = 0$ for $x = \frac 1 4 \pm \frac 3  4$, i.e. for $x = 1$ and for $x = -\frac 1 2$. But $x =1$ is not contained in $(-1,1)$ (in fact $q'(x)$ is not defined for $x = 1$), thus the maximum is attained at $x = -\frac 1 2$.
We therefore get
$$\lvert p(z) \rvert \le 4q(-\frac 1 2) = 3 \sqrt 3 .$$
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