# 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 ${\displaystyle \mathbb {C} }$) 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 preimageEdit

We shall now define and deal with statements of the form

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

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

Definition 2.2.1:

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

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

, we define:

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

Lemma 2.2.2:

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

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

then

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

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

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

, there exists a ${\displaystyle \delta >0}$ such that

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

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

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

, and therefore

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

Lemma 2.2.3:

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

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

, then for all ${\displaystyle B'\subseteq B}$ such that ${\displaystyle z_{0}\in B'}$:

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

Proof:

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

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

Let now ${\displaystyle \epsilon >0}$ be arbitrary. As

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

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

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

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

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

## Continuity of complex functionsEdit

We recall that a function

${\displaystyle f:M\to M'}$

, where ${\displaystyle M}$, ${\displaystyle M'}$ are metric spaces, is continuous if and only if

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

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

Theorem 2.2.4:

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

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

Proof:

## ExercisesEdit

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

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