# 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 preimageEdit

We shall now define and deal with statements of the form

$\lim_{z \to z_0 \atop z \in B'} f(z) = w$

for $B \subseteq \mathbb C$, $f : B \to \mathbb C$, $B' \subseteq B$ and $w \in \mathbb C$, 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

$\forall \epsilon > 0 : \exists \delta > 0 : \left( z \in B' \cap B(z_0, \delta) \Rightarrow |f(z) - w| < \epsilon \right)$

, 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 $\epsilon > 0$ be arbitrary. Since

$\lim_{z \to z_0 \atop z \in B'} f(z) = w$

, there exists a $\delta > 0$ such that

$z \in B' \cap B(z_0, \delta) \Rightarrow |f(z) - w| < \epsilon$

. But since $B'' \subseteq B'$, we also have $B'' \cap B(z_0, \delta) \subseteq B' \cap B(z_0, \delta)$, and thus

$z \in B'' \cap B(z_0, \delta) \Rightarrow z \in B' \cap B(z_0, \delta) \Rightarrow |f(z) - w| < \epsilon$

, and therefore

$\lim_{z \to z_0 \atop z \in B''} f(z) = w$
$////$

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 $\epsilon > 0$ 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:

$z \in B(z_0, \delta) \cap B' \Rightarrow z \in B(z_0, \delta) \Rightarrow z \in B(z_0, \delta_2) \cap U \Rightarrow |f(z) - f(z_0)| < \epsilon$
$////$

## Continuity of complex functionsEdit

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:

## ExercisesEdit

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.

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