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

Example 2.2:

The function

f : C C , f ( z ) := z 2 {\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

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

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

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

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

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

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

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

z B B ( z 0 , δ ) z A B ( z 0 , δ ) | f ( z ) w | < ϵ {\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

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

Proof:

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

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

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

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

there exists a δ 2 > 0 {\displaystyle \delta _{2}>0} such that

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

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

z B ( z 0 , δ ) A z B ( z 0 , δ ) z B ( z 0 , δ 2 ) U | f ( z ) f ( z 0 ) | < ϵ {\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 S C {\displaystyle S\subseteq \mathbb {C} } and f : S C {\displaystyle f:S\to \mathbb {C} } be a function. Then f {\displaystyle f} is defined to be continuous if and only if

z 0 S : lim z z 0 z S f ( z ) = f ( z 0 ) {\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
    f : C C , f ( z ) = { z 2 | z | 2 z 0 1 z = 0 {\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 f {\displaystyle f} is not continuous at 0 {\displaystyle 0} . Hint: Consider the limit with respect to different lines through 0 {\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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