I've seen some posts on this site about this problem, but I didn't understand much of it. I want to show it is differentiable at the origin using only the definition.
$f$ is a complex valued function at $p\in \mathbb{C}$, We say $f$ is differentiable at $p$ if the limit $$\lim_{z\to p} \dfrac{f(z)-f(p)}{z-p}=f^\prime(p)$$ exists
I want to show that for all $\epsilon>0$ there exists $\delta>0$ such that $\bigg|\dfrac{f(z)}{z}\bigg|=\dfrac{|f(z)|}{|z|}<\epsilon$ whenever $|z|<\delta$. So, if we pick $\delta<\epsilon$ then we have $\dfrac{|f(z)|}{|z|}=\dfrac{|z|^2}{|z|}=|z|<\delta<\epsilon$?
Is this the right way of approaching? Thanks!
Yes, this is a good proof, after a couple of minor corrections:
Where you have "... whenever $|z| < \delta$", it would be better to say "... whenever $0 < |z| < \delta$", to explicitly exclude the point $z=0$ where $\left|\frac{f(z)}{z}\right|$ and $\frac{|f(z)|}{|z|}$ have no value at all.
Similarly, "pick $0 < \delta < \epsilon$". Your goal just before mentions that $\delta$ should be positive, but putting that into the choice of $\delta$ makes it clear it happens, and isn't just something you'd want.
Or as pointed out in a comment, it would also work to say just "pick $\delta=\epsilon"$. It doesn't make a whole lot of difference in most $\delta$-$\epsilon$ proofs, but it's stronger in a way to concretely specify an exact $\delta$ (by formula or algorithm) rather than saying one which satisfies the conditions exists.