View source for 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 <math>\mathbb C</math>) and
* characterise continuity of functions mapping from a subset of the complex numbers to the complex numbers using this 'class of limits'.

== Complex functions ==

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Definition 2.1''':

Let <math>S_1, S_2</math> be sets and <math>f: S_1 \to S_2</math> be a function. <math>f</math> is a '''complex function''' if and only if <math>S_1, S_2 \subseteq \mathbb C</math>.}}

'''Example 2.2''':

The function
:<math>f: \mathbb C \to \mathbb C, f(z) := z^2</math>
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
:<math>\lim_{z \to z_0 \atop z \in A} f(z) = w</math>
for <math>S \subseteq \mathbb C</math>, <math>f : S \to \mathbb C</math>, <math>A \subseteq S</math> and <math>w \in \mathbb C</math>, and prove two lemmas about these statements.

{{TextBox| M=0 | W=100% | SG=#FFFFFF |1='''Definition 2.3''':

Let <math>S \subseteq \mathbb C</math> be a set, let <math>f: S \to \mathbb C</math> be a function, let <math>A \subseteq S</math>, let <math>z_0 \in A</math> and let <math>w \in \mathbb C</math>. If
:<math>\forall \epsilon > 0 : \exists \delta > 0 : \left( z \in A \cap B(z_0, \delta) \Rightarrow |f(z) - w| < \epsilon \right)</math>,
we define
:<math>\lim_{z \to z_0 \atop z \in A} f(z) := w</math>.}}

{{TextBox| M=0 | W=100% | SG=#FFFFFF |1='''Lemma 2.4''':

Let <math>S \subseteq \mathbb C</math> be a set, let <math>f: S \to \mathbb C</math> be a function, let <math>B \subseteq A \subseteq S</math>, let <math>z_0 \in B</math> and <math>w \in \mathbb C</math>. If
:<math>\lim_{z \to z_0 \atop z \in A} f(z) = w</math>,
then
:<math>\lim_{z \to z_0 \atop z \in B} f(z) = w</math>.}}

'''Proof''': Let <math>\epsilon > 0</math> be arbitrary. Since
:<math>\lim_{z \to z_0 \atop z \in A} f(z) = w</math>,
there exists a <math>\delta > 0</math> such that
:<math>z \in A \cap B(z_0, \delta) \Rightarrow |f(z) - w| < \epsilon</math>.
But since <math>B \subseteq A</math>, we also have <math>B \cap B(z_0, \delta) \subseteq A \cap B(z_0, \delta)</math>, and thus
:<math>z \in B \cap B(z_0, \delta) \Rightarrow z \in A \cap B(z_0, \delta) \Rightarrow |f(z) - w| < \epsilon</math>,
and therefore
:<math>\lim_{z \to z_0 \atop z \in B} f(z) = w</math>.{{QED}}

{{TextBox| M=0 | W=100% | SG=#FFFFFF |1='''Lemma 2.5''':

Let <math>S \subseteq \mathbb C</math>, <math>f: S \to \mathbb C</math> be a function, <math>O \subseteq S</math> be open, <math>z_0 \in O</math> and <math>w \in \mathbb C</math>. If
:<math>\lim_{z \to z_0 \atop z \in O} f(z) = w</math>,
then for all <math>A \subseteq S</math> such that <math>z_0 \in A</math>
:<math>\lim_{z \to z_0 \atop z \in A} f(z) = w</math>.}}

'''Proof''':

Let <math>A \subseteq S</math> such that <math>z_0 \in A</math>.

First, since <math>O</math> is open, we may choose <math>\delta_1 > 0</math> such that <math>B(z_0, \delta_1) \subseteq O</math>.

Let now <math>\epsilon > 0</math> be arbitrary. As
:<math>\lim_{z \to z_0 \atop z \in O} f(z) = w</math>,
there exists a <math>\delta_2 > 0</math> such that
:<math>z \in B(z_0, \delta_2) \cap U \Rightarrow |f(z) - f(z_0)| < \epsilon</math>.

We define <math>\delta := \min\{\delta_1, \delta_2\}</math> and obtain
:<math>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</math>.{{QED}}

== Continuity of complex functions ==

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Definition 2.6''':

Let <math>S \subseteq \mathbb C</math> and <math>f: S \to \mathbb C</math> be a function. Then <math>f</math> is defined to be continuous if and only if
:<math>\forall z_0 \in S : \lim_{z \to z_0 \atop z \in S} f(z) = f(z_0)</math>.}}

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