# Complex Analysis/Complex Functions/Continuous 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'.

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

We shall now define and deal with statements of the form

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

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

Definition 2.2.1:

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

$\forall \varepsilon >0:\exists \delta >0:{\bigl (}z\in B'\cap B(z_{0},\delta )\Rightarrow |f(z)-w|<\varepsilon {\bigr )}$

we define:

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

Lemma 2.2.2:

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

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

then

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

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

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

there exists a $\delta >0$  such that

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

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

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

and therefore

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

Lemma 2.2.3:

Let $B\subseteq \mathbb {C}$  , $f:B\to \mathbb {C}$  be a function, $O\subseteq B$  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 $B'\subseteq B$  such that $z_{0}\in B'$  :

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

Let $B'\subseteq B$  such that $z_{0}\in B'$  .

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

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

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

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

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

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

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

## Continuity of complex functions

We recall that a function

$f:M\to M'$

where $M,M'$  are metric spaces, is continuous if and only if

$x_{l}\to x,l\to \infty \Rightarrow f(x_{l})\to f(x)$

for all convergent sequences $(x_{l})_{l\in \mathbb {N} }$  in $M$  .

Theorem 2.2.4:

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

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

## Exercises

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