A complex inequality

Let $z \in \mathbb{C}$ such that $|z+1| \leq 1$ and $|z^2+1|\leq 1$ . Prove that $|z|\leq 1$ .

Solution

We have successively:

$\begin{align*} \left | 2z \right | &= \left | \left ( z +1 \right )^2 - \left ( z^2+1 \right ) \right | \\ &\leq \left | z +1 \right |^2 + \left | z^2+1 \right | \\ &=1^2 + 1 \\ &=2 \end{align*}$

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