$f: \mathbb{C} \rightarrow \mathbb{C}, \; z \mapsto f(z) := \frac{z^2}{1+|z|}$
How would I go about proving this is uniformly continuous? Currently I have only practically dealt with analyzing continuity of real functions. Here's my approach:
$\forall \epsilon > 0 \; \exists \delta > 0 \; \forall z,z_0 \in \mathbb{C}: |z-z_0| < \delta \implies |f(z) - f(z_0)| < \epsilon$
Let $\epsilon > 0$ and $|z-z_0| < \delta$. We don't know yet whether $\delta$ exists or its value. Looking at $|f(z) - f(z_0)|$ we get $|\frac{z^2}{1+|z|} - \frac{z_0^2}{1+|z_0|}|$. Now I know that I will need to estimate this term somehow to eliminate $z_0$ such that our $\delta$ is only dependent on $\epsilon$.
First thing I thought about was bringing both summands down to a common denominator such that we have $$\frac{|z^2(1+|z_0|) - z_0^2(1+|z|)|}{|(1+|z|) (1+|z_0|)|}$$
So now when we multiply this out we get $$ frac{z^2 - z_0^2 + z^2|z_0| - z_0^2|z|}{(1+|z|) (1+|z_0|)} < \frac{\delta + z^2|z_0| - z_0^2|z|}{(1+|z|) (1+|z_0|)} $$
My lectures have just introduced this topic and I don't really know yet how to go about these proofs. Thanks for any help!