Limits of Complex Functions

Complex Limits

Definition: Limit of a Complex Function

Let $f:\mathcal{D}\subseteq \mathbb{C}\to \mathbb{C}$ be a complex function and let $c\in \mathbb{C}$ be an accumulation point of $\mathcal{D}$.

We say that $L$ is the limit of $f$ as $z$ approaches $c$ if for each $\epsilon >0$ there exists some $\delta >0$ such that

$$0<|z-c|<\delta ⟹|f(z)-L|<\epsilon$$

for all $z\in \mathcal{D}$.

NOTATION

Most commonly, the limit is denoted by

$$\underset{z\to c}{\lim }f(z)=L$$

In text, one also writes ”$f(z)\to L$ as $z\to c$“. Sometimes, one might also encounter $f(z)\underset{z\to c}{⟶}L$ and $f(z)\stackrel{z\to c}{⟶}L$.

Definition: Limit at Infinity

Let $f:\mathcal{D}\subseteq \mathbb{C}\to \mathbb{C}$ be a complex function such that $\mathcal{D}$ is the complement of some open ball in $\mathbb{C}$ centered at zero.

We say that $L\in \mathbb{C}$ is the limit of $f$ for $z\to \infty$ if for each $\epsilon >0$ there exists some $R>0$ such that

$$|z|>R⟹|f(z)-L|<\epsilon$$

for all $z\in \mathcal{D}$.

NOTATION

$$\underset{z\to \infty }{\lim }f(z)=L$$

Theorem: Uniqueness of the Limit

Let $f:\mathcal{D}\subseteq \mathbb{C}\to \mathbb{C}$ be a complex function and let $c\in \mathbb{C}$ be an accumulation point of $\mathcal{D}$ or $c=\infty$.

If the limit of $f$ exists at $c$, then it is unique:

$$\underset{z\to c}{\lim }f(z)=L\in \mathbb{C}\text{ and }\underset{z\to c}{\lim }f(z)=M\in \mathbb{C}⟹L=M$$

PROOF

TODO

Theorem: Complex Limits via Absolute Value

Let $f:\mathcal{D}\subseteq \mathbb{C}\to \mathbb{C}$ be a complex function and let $c\in \mathbb{C}$ be an accumulation point of $\mathcal{D}$.

A number $L\in \mathbb{C}$ is the limit of $f$ for $z\to c$ if and only if

$$\underset{z\to c}{\lim }|f(z)-L|=0$$

PROOF

TODO

Theorem: Component-wise Limits

Let $f:\mathcal{D}\subseteq \mathbb{C}\to \mathbb{C}$ be a complex function and let $c\in \mathbb{C}$ be an accumulation point of $\mathcal{D}$.

The limit of $f$ at $c$ is $L\in \mathbb{C}$ if and only if the limit of $f$‘s real part is $\mathrm{Re}(L)$ and the limit of its imaginary part if $\mathrm{Im}(L)$:

$$\underset{z\to c}{\lim }f(z)=L⟺\underset{z\to c}{\lim }\mathrm{Re}f(z)=\mathrm{Re}(L)\text{and}\underset{z\to c}{\lim }\mathrm{Im}f(z)=\mathrm{Im}(L)$$

PROOF

TODO

Theorem: Operations with Limits

Let $f:{\mathcal{D}}_f\subseteq \mathbb{C}\to \mathbb{C}$ and $g:{\mathcal{D}}_g\subseteq \mathbb{C}\to \mathbb{C}$ be complex functions and let $c\in \mathbb{C}$ be an accumulation point of ${\mathcal{D}}_f\cap {\mathcal{D}}_g$.

If the limits of $f$ and $g$ exist at $c$, then

$$\underset{z\to c}{\lim }\left(\alpha f(z)+\beta g(z)\right)=\alpha \underset{z\to c}{\lim }f(z)+\beta \underset{z\to c}{\lim }g(z)\forall \alpha ,\beta \in \mathbb{C}$$

Moreover, if there also exists some open ball around $c$ on which $f$ and $g$ are bounded, then

$$\begin{array}{rl}& \underset{z\to c}{\lim }\left(f(z)g(z)\right)=\left(\underset{z\to c}{\lim }f(z)\right)\cdot \left(\underset{z\to c}{\lim }g(z)\right)\\ & \\ & \underset{z\to c}{\lim }\frac{f(z)}{g(z)}=\frac{\underset{z\to c}{\lim }f(z)}{\underset{z\to c}{\lim }g(z)},\text{ provided that }\underset{z\to c}{\lim }g(z)\ne 0\end{array}$$

PROOF

TODO

Squeeze Theorem

Let $f:{\mathcal{D}}_f\subseteq \mathbb{C}\to \mathbb{C}$ and $g:{\mathcal{D}}_g\subseteq \mathbb{C}\to \mathbb{C}$ be complex functions and let $c\in \mathbb{C}$ be an accumulation point of ${\mathcal{D}}_f\cap {\mathcal{D}}_g$.

If there exists some deleted neighborhood $N$ of $c$ such that $|g(z)|\le |f(z)|$ for all $z\in N$ and ${\lim }_{z\to c}f(z)=0$, then ${\lim }_{z\to c}g(z)=0$.

If there exists some deleted neighborhood $N$ of $c$ on which $g$ is bounded and ${\lim }_{z\to c}f(z)=0$, then ${\lim }_{z\to c}(f(z)\cdot g(z))=0$.

PROOF

TODO

Theorem

Let $f:\mathcal{D}\subseteq \mathbb{C}\to \mathbb{C}$ be a complex function such that $\mathcal{D}$ is the complement of some open ball in $\mathbb{C}$ centered at zero.

The limit of $f$ for $z\to \infty$ is $L\in \mathbb{C}$ if and only if the limit of $|f(z)-L|$ for $|z|\to \infty$ is zero.

$$\underset{z\to \infty }{\lim }f(z)=L⟺\underset{|z|\to \infty }{\lim }|f(z)-L|=0$$

PROOF

TODO

Theorem: Limit $\leftrightarrow$ Limit at Infinity

Let $f:\mathcal{D}\subseteq \mathbb{C}\to \mathbb{C}$ be a complex function such that $\mathcal{D}$ is the complement of some open ball in $\mathbb{C}$ centered at zero.

The limit of $f$ for $z\to \infty$ is equal to $L\in \mathbb{C}$ if and only if the limit of $f\left(\frac{1}{z}\right)$ for $z\to 0$ is $L$.

$$\underset{z\to \infty }{\lim }f(z)=L⟺\underset{z\to 0}{\lim }f\left(\frac{1}{z}\right)=L$$

PROOF

TODO

Infinite Limits

Definition: Infinite Limits

Let $c\in \mathbb{C}$ and let $f:\mathcal{D}\subseteq \mathbb{C}\to \mathbb{C}$ be a complex function defined on some deleted neighborhood $\mathcal{D}$ of $c$.

We say that $f$ has an infinite limit at $c$ if for each $M>0$, there exists some $\delta >0$ such that

$$0<|z-c|<\delta ⟹|f(z)|>M$$

for all $z\in \mathcal{D}$.

NOTATION

$$\underset{z\to c}{\lim }f(z)=\infty$$

Definition: Infinite Limits at Infinity

Let $f:\mathcal{D}\subseteq \mathbb{C}\to \mathbb{C}$ be a complex function such that $\mathcal{D}$ is the complement of some open ball in $\mathbb{C}$ centered at zero.

We say that $f$ has an infinite limit for $z\to \infty$ if for each $M>0$ there exists some $R>0$ such that

$$|z|>R⟹|f(z)|>M$$

for all $z\in \mathcal{D}$.

NOTATION

$$\underset{z\to \infty }{\lim }f(z)=\infty$$

WARNING

Even though we use limit notation for infinite limits and infinite limits at infinity, we never say that these limits exist, since they are not complex numbers.

Theorem

Let $c\in \mathbb{C}$ and let $f:\mathcal{D}\subseteq \mathbb{C}\to \mathbb{C}$ be a complex function defined on some deleted neighborhood $\mathcal{D}$ of $c$.

The limit of $f$ for $z\to c$ is $\infty$ if and only if the limit of $|f|$ for $z\to c$ is $\infty$.

$$\underset{z\to c}{\lim }f(z)=\infty ⟺\underset{z\to c}{\lim }|f(z)|=\infty$$

PROOF

TODO

Theorem

Let $f:\mathcal{D}\subseteq \mathbb{C}\to \mathbb{C}$ be a complex function such that $\mathcal{D}$ is the complement of some open ball in $\mathbb{C}$ centered at zero.

The limit of $f$ for $z\to \infty$ is $\infty$ if and only if the limit of $|f|$ for $|z|\to \infty$ is $\infty$.

$$\underset{z\to \infty }{\lim }f(z)=\infty ⟺\underset{|z|\to \infty }{\lim }|f(z)|=\infty$$

PROOF

TODO

Theorem: Infinite Limit $\leftrightarrow$ Infinite Limit at Infinity

Let $f:\mathcal{D}\subseteq \mathbb{C}\to \mathbb{C}$ be a complex function such that $\mathcal{D}$ is the complement of some open ball in $\mathbb{C}$ centered at zero.

The limit of $f$ for $z\to \infty$ is $\infty$ if and only if the limit of $f\left(\frac{1}{z}\right)$ for $z\to 0$ is $\infty$.

$$\underset{z\to \infty }{\lim }f(z)=\infty ⟺\underset{z\to 0}{\lim }f\left(\frac{1}{z}\right)=\infty$$

PROOF

TODO

Theorem: Common Limits

Following are some limits for complex functions:

$$\begin{array}{rl}& \underset{z\to c}{\lim }\lambda =\lambda \lambda ,c\in \mathbb{C}\\ & \underset{z\to \infty }{\lim }\frac{1}{z^n}=0\forall n\in \mathbb{N}\end{array}$$

PROOF

TODO

Sources

1. N. H. Asmar, L. Grafakos, “Analytic Functions,” in Complex Analysis with Applications, Columbia, USA: Springer, 2018