$e^{1/z}$ has an essential singularity at $z=0$, which by the Casorati–Weierstrass theorem implies that for every complex number $W$ there is a sequence $z_k\to 0$ with $e^{1/z_k}\to W$.
Is there some elegant way to see this directly without proving Casorati–Weierstrass first? (I know that the proof of Casorati–Weierstrass is not hard, I'm just curious.)
Best Answer
Let $W = r e^{\theta i} = e^{\ln r + \theta i}$. Then it is also true that for any integer $k$, $W = e^{\ln r + (2\pi k + \theta)i}$.
Let $z_k = \frac1{\ln r + (2\pi k+\theta)i}$; then $e^{1/z_k}$ exactly equals $W$ for all $k$, but as $k \to \infty$, $z_k \to 0$ because $|\ln r + (2\pi k+\theta)i| \to \infty$.
(As pointed out by J.G. in the comments, if $W=0$ this will not work because we cannot take $\ln 0$, but we can treat $W=0$ as a special case, defining $z_k = -\frac1k$.)
By the way, the fact that we can achieve $e^{1/z_k} = W$ exactly rather than in the limit, provided $W\ne 0$, is an instance of Picard's theorem.