Complex Analysis/Example - exp(1/z)

Introduction

We investigate sequences approaching $0\in \mathbb {C}$ 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 $w\in \mathbb {C}$ and any punctured $\varepsilon$-neighborhood around 0, there exists a sequence $\left(z^{(n)}\right){n\in \mathbb {N} }$ such that the image sequence $\left(f(z^{(n)})\right){n\in \mathbb {N} }$ converges to $w\in \mathbb {C}$.

Laurent Series for exp(1/z)

First, we note the Laurent series for $f(z)=e^{\frac {1}{z}}$ with $z=z_{1}+iz_{2}\in \mathbb {C} \setminus {0}$ using the definition of the Taylor series expanded at the point $z_{0}=0$: $f(z)=e^{\frac {1}{z}}=e^{z^{-1}}=\displaystyle \sum _{n=0}^{+\infty }{\frac {z^{-n}}{n!}}=\displaystyle \sum _{n=-\infty }^{0}{\frac {z^{n}}{(-n)!}}=\displaystyle \sum _{n=-\infty }^{0}{\frac {1}{(-n)!}}\cdot z^{n}$.

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

Image Points of Punctured $\varepsilon$-Neighborhoods

As a special case of the Casorati-Weierstrass theorem, we constructively demonstrate for $f(x):=e^{\frac {1}{z}}$: $\forall _{w\in \mathbb {C} }\exists {\left(z^{(n)}\right){n\in \mathbb {N} }}$ such that $\displaystyle \lim {n\to \infty }z^{(n)}=0$ $\displaystyle \lim _{n\to \infty }f(z^{(n)})=w=w_{1}+iw_{2}$.

Proof (Constructive)

For the image points $w=w_{1}+iw_{2}\in \mathbb {C}$, we distinguish two cases:

Case 1: $w\neq 0$

Case 2: $w=0$

Case 1:

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

Sequence Definition (Case 1)

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

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

Case 2:

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

Sequence Definition (Case 2)

Using the property of the exponential function in $\mathbb {R}$ with $w=0$: $\displaystyle \lim _{n\to \infty }e^{-n}=0$ $z^{(n)}=z_{1}^{(n)}+iz_{2}^{(n)}=-{\frac {1}{n}}+i\cdot 0\in \mathbb {C} \setminus {0}$.

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

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• Date: 12/30/2024