Fix \(a\in \CC\union\ts{\infty}\) and let \(f(z) \da e^{1\over z^2}\). Find a sequence \(z_k\to 0\) such that \(f(z_k) \convergesto{k\to\infty} a\)
Solution
If \(a\in\CC^\times\), choose any logarithm \(L\in\CC\) with \(e^L=a\) and set \[ z_k=\qty{L+2\pi i k}^{-1/2}, \] choosing either square root for each \(k\). Then \(z_k\to0\) and \[ {1\over z_k^2}=L+2\pi i k, \qquad f(z_k)=e^{L+2\pi i k}=a. \]
For \(a=0\), take \(z_k=i/\sqrt{k}\). Then \(z_k\to0\) and \[ f(z_k)=e^{-k}\to0. \]
For \(a=\infty\), take \(z_k=1/\sqrt{k}\). Then \(z_k\to0\) and \[ f(z_k)=e^k\to\infty \] in \(\CP^1\).