Given \(w\in\mathbb C\), find an explicit sequence \(z_n\to0\) such that \[ e^{1/z_n}\to w. \]
Solution
If \(w\neq0\), choose any logarithm \(L\in\mathbb C\) with \(e^L=w\) and set \[ z_n=\frac{1}{L+2\pi i n},\qquad n\ge1. \] Then \(z_n\to0\) and in fact \(e^{1/z_n}=w\) for every \(n\).
If \(w=0\), take \[ z_n=-\frac1n. \] Then \(z_n\to0\) and \(e^{1/z_n}=e^{-n}\to0\).
1. Suppose \(w\neq0\). There is \(L\in\mathbb C\) such that \(e^L=w\).
Write \(w=re^{i\theta}\) with \(r>0\). Then \[ L=\log r+i\theta \] satisfies \(e^L=w\).
2. For\[ z_n=\frac{1}{L+2\pi i n}, \]one has \(z_n\to0\) and \(e^{1/z_n}=w\) for every \(n\).
Since \[ |L+2\pi i n|\longrightarrow\infty, \] we have \(z_n\to0\). Moreover \[ e^{1/z_n} =e^{L+2\pi i n} =e^L e^{2\pi i n} =w. \] Thus the required convergence is exact term-by-term.
3. If \(w=0\), the sequence \(z_n=-1/n\) has the required property.
Clearly \(z_n\to0\), while \[ e^{1/z_n}=e^{-n}\longrightarrow0=w. \]