Prove or disprove the uniform continuity of $f: \mathbb{C} \to \mathbb{C}, z \mapsto f(z)=\frac{z^2}{1+|z|}$

Question

I tried using the definition of the continuity to try and find a $\delta$ but I got stuck here
$$ |f(w) - f(z)| = |\frac{w^2 \cdot (1+|z|) - z^2 \cdot (1+|w|)}{(1+|w|)\cdot(1+|z|)}|$$
What can I do to find a suitable $\delta$ that would end my proof?

Answer 1

Without loss of generality, $|z| \le |w|$. Then $$ f(w) - f(z) = \frac{w^2-z^2}{1+|w|} + z^2 \left( \frac{1}{1+|w|} - \frac{1}{1+|z|} \right) \\ = \frac{w+z}{1+|w|} (w-z) + \frac{z^2}{(1+|w|)(1+|z|)} (|z| - |w|) $$ and therefore $$ \begin{align} |f(w) - f(z)| &\le \frac{|w|+|z|}{1+|w|} |w-z| + \frac{|z|^2}{(1+|w|)(1+|z|)} |w-z| \\ &\le \frac{2|w|}{1+|w|} |w-z|+ \frac{|w||z|}{(1+|w|)(1+|z|)} |w-z| \\ &\le 3 |w-z| \end{align} $$ so that $f$ is Lipschitz continuous (and in particular uniformly continuous).