I need to prove that the following function is continuous at all points:
$f:\mathbb{C} \rightarrow \mathbb{C} \hspace{0.3cm} \text{ given by } \hspace{0.3cm} f(z)=\frac{2|z|}{i+\text{Re}z}$
I understand that to show continuity at a point $w$ I must show that ,
$(\forall \epsilon > 0)(\exists \delta > 0)\text{ Such that }|f(z)-f(w)| \leq \epsilon\text{ whenever }z \in \mathbb{C}\text{ is such that }|z-w| < \delta$
I'm not sure how to go about finding such a $\delta$ and would appreciate a point in the right direction.
Sometimes these things are easier if you abstract them a little, the details often add clutter.
Let $g(z) = 2z$ and $h(z) = i+ \operatorname{re} z$. Note that $|h(z)| \ge 1$ and $f = {g \over h}$.
$|f(z)-f(w)| = { |g(z)h(w) - g(w)h(z)| \over |h(z)| |h(w)| } \le |g(z)h(w) - g(w)h(z)|$.
Now note that we can estimate this (add & subtract $g(w)h(w)$) to get $|f(z)-f(w)| \le |g(w) |h(z)-h(w)| + |h(w)||g(z)-g(w)|$.
Now note that $g(z)-g(w) = 2 (z-w)$ and $h(z)-h(w) = \operatorname{re}(z-w)$. (Also note that $|\operatorname{re} z | \le |z|$.)
So we get $|f(z)-f(w)| \le 2|w||z-w|+2|i+\operatorname{re} w||z-w| = 2(|w|+ |i+\operatorname{re} w| ) |z-w|$.
Note that the quantity $K=2(|w|+ |i+\operatorname{re} w| )$ is a fixed number (we are looking at continuity around $w$), so we have $|f(z)-f(w)| \le K |z-w|$.
So, given $\epsilon>0$ choose $\delta = {\epsilon \over K}$ and then if $|z-w| < \delta$ we will have $|f(z)-f(w)| < \epsilon$.