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 $\varepsilon \gt 0$ there exists some $\delta \gt 0$ such that

\[ 0 \lt |z - c| \lt \delta \implies |f(z) - L| \lt \varepsilon \]

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

Notation

Most commonly, the limit is denoted by

\[ \lim_{z \to c} f(z) = L \qquad \]

In text, one also writes "$f(z) \to L$ as $z \to c$". Sometimes, one might also encounter $f(z) \underset{z \to c}{\longrightarrow} L$ and $f(z) \overset{z \to c}{\longrightarrow} L$.

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 $\varepsilon \gt 0$ there exists some $R \gt 0$ such that

\[ |z| \gt R \implies |f(z) - L| \lt \varepsilon \]

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

Notation

\[ \lim_{z \to \infty} f(z) = L \]

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:

\[ \lim_{z \to c} f(z) = L \in \mathbb{C} \qquad \text{ and } \qquad \lim_{z \to c} f(z) = M \in \mathbb{C} \implies L = M \]

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

\[ \lim_{z \to c} |f(z) - L| = 0 \]

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:

\[ \lim_{z \to c} f(z) = L \iff \lim_{z \to c} \operatorname{Re} f (z) = \operatorname{Re} (L) \qquad \text{and} \qquad \lim_{z \to c}\operatorname{Im} f (z) = \operatorname{Im} (L) \]

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

\[ \lim_{z\to c} \left( \alpha f(z) + \beta g(z) \right) = \alpha \lim_{z \to c} f(z) + \beta \lim_{z \to c} g(z) \qquad \forall \alpha, \beta \in \mathbb{C} \]

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

$$
\begin{aligned}

&\lim_{z \to c} \left(f(z) g(z)\right) = \left(\lim_{z \to c} f(z)\right) \cdot \left(\lim_{z \to c} g(z)\right) \

\

&\lim_{z \to c} \frac{f(z)}{g(z)} = \frac{\displaystyle \lim_{z \to c} f(z)}{\displaystyle \lim_{z \to c} g(z)}, \qquad \text{ provided that } \lim_{z \to c} g(z) \ne 0

\end{aligned}
$$

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.

\[ \lim_{z \to \infty} f(z) = L \iff \lim_{|z| \to \infty} |f(z) - L| = 0 \]

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 $L$.

\[ \lim_{z \to \infty} f(z) = L \iff \lim_{z \to 0} f \left( \frac{1}{z} \right) = L \]

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 $M \gt 0$, there exists some $\delta \gt 0$ such that

\[ 0 \lt |z - c| \lt \delta \implies |f(z)| \gt M \]

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

Notation

\[ \lim_{z \to c} f(z) = \infty \]

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 $M \gt 0$ there exists some $R \gt 0$ such that

\[ |z| \gt R \implies |f(z)| \gt M \]

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

Notation

\[ \lim_{z \to \infty} f(z) = \infty \]

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](../../Real%20Analysis/Real%20Functions/Limits%20(Real%20Functions.md#Infinite%20Limits) of $|f|$ for $z \to c$ is $\infty$.

\[ \lim_{z \to c} f(z) = \infty \iff \lim_{z \to c} |f(z)| = \infty \]

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 $\infty$.

\[ \lim_{z \to \infty} f(z) = \infty \iff \lim_{|z| \to \infty } |f(z)| = \infty \]

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 $\infty$.

\[ \lim_{z \to \infty} f(z) = \infty \iff \lim_{z \to 0} f\left( \frac{1}{z} \right) = \infty \]

Proof

TODO

Theorem: Common Limits

Following are some limits for complex functions:

$$
\begin{aligned}

&\lim_{z \to c} \lambda = \lambda \qquad \lambda, c \in \mathbb{C} \

&\lim_{z \to \infty} \frac{1}{z^n} = 0 \qquad \forall n \in \mathbb{N}

\end{aligned}
$$

Proof

TODO

Sources#

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