How to Find Continuity of Complex Function

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'.

Complex functions [edit | edit source]

Definition 2.1:

Let ${\displaystyle S_{1},S_{2}}$ be sets and ${\displaystyle f:S_{1}\to S_{2}}$ be a function. ${\displaystyle f}$ is a complex function if and only if ${\displaystyle S_{1},S_{2}\subseteq \mathbb {C} }$ .

Example 2.2:

The function

${\displaystyle f:\mathbb {C} \to \mathbb {C} ,f(z):=z^{2}}$

is a complex function.

Limits of complex functions with respect to subsets of the preimage [edit | edit source]

We shall now define and deal with statements of the form

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

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

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

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

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

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

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

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

and therefore

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

Proof:

Let ${\displaystyle A\subseteq S}$ such that ${\displaystyle z_{0}\in A}$ .

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

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

Continuity of complex functions [edit | edit source]

Definition 2.6:

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

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

Exercises [edit | edit source]

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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Source: https://en.wikibooks.org/wiki/Complex_Analysis/Limits_and_continuity_of_complex_functions

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