Complex Analysis/Example - exp(1/z)

Introduction

We investigate sequences approaching $0\in ℂ$ and the behavior of $f(z)=e^{\frac{1}{z}}$ for these sequences converging to the essential singularity at 0. This constructive approach demonstrates that for any image point $w\in ℂ$ and any punctured $\epsilon$-neighborhood around 0, there exists a sequence $\left(z^{(n)}\right)n\in \mathbb{N}$ such that the image sequence $(f(z^{(n)}))n\in \mathbb{N}$ converges to $w\in ℂ$.

Laurent Series for exp(1/z)

First, we note the Laurent series for $f(z)=e^{\frac{1}{z}}$ with $z=z_1+i{z}_2\in ℂ\setminus 0$ using the definition of the Taylor series expanded at the point $z_0=0$: $f(z)=e^{\frac{1}{z}}=e^{z^{-1}}=\sum _{n=0}^{+\mathrm{\infty }}\frac{z^{-n}}{n!}=\sum _{n=-\mathrm{\infty }}^0\frac{z^n}{(-n)!}=\sum _{n=-\mathrm{\infty }}^0\frac{1}{(-n)!}\cdot z^n$.

Now, compute the Laurent expansion of $f(z)=e^{\frac{1}{z}}$ with an expansion point $z_0\ne 0$.

Image Points of Punctured $\epsilon$-Neighborhoods

As a special case of the Casorati-Weierstrass theorem, we constructively demonstrate for $f(x):=e^{\frac{1}{z}}$: ${\mathrm{\forall }}_{w\in ℂ}\mathrm{\exists }\left(z^{(n)}\right)n\in \mathbb{N}$ such that $\lim n\to \mathrm{\infty }{z}^{(n)}=0$ $\underset{n\to \mathrm{\infty }}{\lim }f(z^{(n)})=w=w_1+i{w}_2$.

Proof (Constructive)

For the image points $w=w_1+i{w}_2\in ℂ$, we distinguish two cases:

Case 1: $w\ne 0$

Case 2: $w=0$

Case 1:

Let $w=w_1+i{w}_2\in ℂ\setminus 0$ be arbitrarily chosen. Define a sequence ${\left(z^{(n)}\right)}_{n\in \mathbb{N}}$ in $z^{(n)}=z^{(n)}1+i{z}^{(n)}2\in ℂ\setminus 0$ such that $\lim n\to \mathrm{\infty }{z}^{(n)}=0$ $\lim n\to \mathrm{\infty }f(z^{(n)})=w=w_1+i{w}_2$.

Sequence Definition (Case 1)

We use the polar representation of $w=w_1+i{w}_2\in ℂ\setminus 0$: $w=w_1+i{w}_2=|w|\cdot (\cos (\alpha )+i\cdot \sin (\alpha ))=|w|\cdot e^{i\alpha }$ $z^{(n)}=z_1^{(n)}+i{z}_2^{(n)}=\frac{1}{\log (|w|)+i\cdot (\alpha +2\pi n)}\in ℂ\setminus 0$.

We demonstrate the convergence property: $\underset{n\to \mathrm{\infty }}{\lim }f\left(z^{(n)}\right)=\underset{n\to \mathrm{\infty }}{\lim }f\left(\frac{1}{\log (|w|)+i\cdot (\alpha +2\pi n)}\right)=\underset{n\to \mathrm{\infty }}{\lim }{e}^{\log (|w|)+i\cdot (\alpha +2\pi n)}$ $=\underset{n\to \mathrm{\infty }}{\lim }|w|\cdot e^{i\alpha +2\pi n}=|w|\cdot e^{i\alpha }=w=w_1+i{w}_2$.

Case 2:

Let $w=0$. Define a sequence ${\left(z^{(n)}\right)}_{n\in \mathbb{N}}$ in $z^{(n)}=z^{(n)}1+i{z}^{(n)}2\in ℂ\setminus 0$ such that $\lim n\to \mathrm{\infty }{z}^{(n)}=0$ $\lim n\to \mathrm{\infty }f(z^{(n)})=0$.

Sequence Definition (Case 2)

Using the property of the exponential function in $ℝ$ with $w=0$: $\underset{n\to \mathrm{\infty }}{\lim }{e}^{-n}=0$ $z^{(n)}=z_1^{(n)}+i{z}_2^{(n)}=-\frac{1}{n}+i\cdot 0\in ℂ\setminus 0$.

Now, we demonstrate the convergence properties: $\underset{n\to \mathrm{\infty }}{\lim }{z}^{(n)}=\underset{n\to \mathrm{\infty }}{\lim }-\frac{1}{n}=0$ $\underset{n\to \mathrm{\infty }}{\lim }f\left(z^{(n)}\right)=\underset{n\to \mathrm{\infty }}{\lim }f(-\frac{1}{n})=\underset{n\to \mathrm{\infty }}{\lim }{e}^{-n}=0$.

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• Date: 12/30/2024

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