continuity of $e^{-1/z}$ in $\mathbb C$

Question

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?

Answer 1

For continuity, note that $-1/z$ maps $\{0<|z|<1\}$ into $\{1<|z|<\infty\}$ continuously, $e^z$ is continuous everywhere, and the composition of continuous maps is continuous. As for uniform continuity, consider the behavior of $f(x+i\sqrt x) - f(x)$ as $x\to 0^+.$

Answer 2

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.

Answer 3

As $f(z)=-\frac{1}{z}$ is analytic in $0<\vert z\vert<1$ then so is $e^{-\frac{1}{z}}$.