# Complex Analysis/Limits and continuity of complex functions

In this section, we

• introduce a 'broader class of limits' than known from real analysis (namely limits with respect to a subset of $\mathbb {C}$ ) and
• characterise continuity of functions mapping from a subset of the complex numbers to the complex numbers using this 'class of limits'.

## Complex functions

Definition 2.1:

Let $S_{1},S_{2}$  be sets and $f:S_{1}\to S_{2}$  be a function. $f$  is a complex function if and only if $S_{1},S_{2}\subseteq \mathbb {C}$ .

Example 2.2:

The function

$f:\mathbb {C} \to \mathbb {C} ,f(z):=z^{2}$

is a complex function.

## Limits of complex functions with respect to subsets of the preimage

We shall now define and deal with statements of the form

$\lim _{z\to z_{0} \atop z\in A}f(z)=w$

for $S\subseteq \mathbb {C}$ , $f:S\to \mathbb {C}$ , $A\subseteq S$  and $w\in \mathbb {C}$ , and prove two lemmas about these statements.

Definition 2.3:

Let $S\subseteq \mathbb {C}$  be a set, let $f:S\to \mathbb {C}$  be a function, let $A\subseteq S$ , let $z_{0}\in A$  and let $w\in \mathbb {C}$ . If

$\forall \epsilon >0:\exists \delta >0:\left(z\in A\cap B(z_{0},\delta )\Rightarrow |f(z)-w|<\epsilon \right)$ ,

we define

$\lim _{z\to z_{0} \atop z\in A}f(z):=w$ .

Lemma 2.4:

Let $S\subseteq \mathbb {C}$  be a set, let $f:S\to \mathbb {C}$  be a function, let $B\subseteq A\subseteq S$ , let $z_{0}\in B$  and $w\in \mathbb {C}$ . If

$\lim _{z\to z_{0} \atop z\in A}f(z)=w$ ,

then

$\lim _{z\to z_{0} \atop z\in B}f(z)=w$ .

Proof: Let $\epsilon >0$  be arbitrary. Since

$\lim _{z\to z_{0} \atop z\in A}f(z)=w$ ,

there exists a $\delta >0$  such that

$z\in A\cap B(z_{0},\delta )\Rightarrow |f(z)-w|<\epsilon$ .

But since $B\subseteq A$ , we also have $B\cap B(z_{0},\delta )\subseteq A\cap B(z_{0},\delta )$ , and thus

$z\in B\cap B(z_{0},\delta )\Rightarrow z\in A\cap B(z_{0},\delta )\Rightarrow |f(z)-w|<\epsilon$ ,

and therefore

$\lim _{z\to z_{0} \atop z\in B}f(z)=w$ .$\Box$

Lemma 2.5:

Let $S\subseteq \mathbb {C}$ , $f:S\to \mathbb {C}$  be a function, $O\subseteq S$  be open, $z_{0}\in O$  and $w\in \mathbb {C}$ . If

$\lim _{z\to z_{0} \atop z\in O}f(z)=w$ ,

then for all $A\subseteq S$  such that $z_{0}\in A$

$\lim _{z\to z_{0} \atop z\in A}f(z)=w$ .

Proof:

Let $A\subseteq S$  such that $z_{0}\in A$ .

First, since $O$  is open, we may choose $\delta _{1}>0$  such that $B(z_{0},\delta _{1})\subseteq O$ .

Let now $\epsilon >0$  be arbitrary. As

$\lim _{z\to z_{0} \atop z\in O}f(z)=w$ ,

there exists a $\delta _{2}>0$  such that

$z\in B(z_{0},\delta _{2})\cap U\Rightarrow |f(z)-f(z_{0})|<\epsilon$ .

We define $\delta :=\min\{\delta _{1},\delta _{2}\}$  and obtain

$z\in B(z_{0},\delta )\cap A\Rightarrow z\in B(z_{0},\delta )\Rightarrow z\in B(z_{0},\delta _{2})\cap U\Rightarrow |f(z)-f(z_{0})|<\epsilon$ .$\Box$

## Continuity of complex functions

Definition 2.6:

Let $S\subseteq \mathbb {C}$  and $f:S\to \mathbb {C}$  be a function. Then $f$  is defined to be continuous if and only if

$\forall z_{0}\in S:\lim _{z\to z_{0} \atop z\in S}f(z)=f(z_{0})$ .

## Exercises

1. Prove that if we define
$f:\mathbb {C} \to \mathbb {C} ,f(z)={\begin{cases}{\frac {z^{2}}{|z|^{2}}}&z\neq 0\\1&z=0\end{cases}}$ ,
then $f$  is not continuous at $0$ . Hint: Consider the limit with respect to different lines through $0$  and use theorem 2.2.4.