# 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

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