continuity of $e^{-1/z}$ in $\mathbb C$ Let f be the function defined by $f(z)=e^{-1/z}$ in $\mathbb C$, prove that $f$ is continuous in the set $0< \vert z \vert < 1$ and $\vert arg(z) \vert <\pi/2$ but it's not uniformly continuous on it.
I think an easy way to prove it is firstly to show that $f$ is analytic, therefore f is differentiable. Hence $f$ is continuous. 
Is there an easier way to prove it?

Sure. Pick any $z \neq 0.$ Then let $B$ be a ball of radius $r = |z|/2$. Restricted to just on that ball, the function $g(z)=-1/z$ is continuous. Also exp is continuous everywhere. Thus $f=exp(g)$ is the composite of continuous functions. 
