Let
G
⊆
C
{\displaystyle G\subseteq \mathbb {C} }
z
0
∈
G
{\displaystyle z_{0}\in G}
f
:
G
∖
z
0
→
C
{\displaystyle f\colon G\setminus {z_{0}}\to \mathbb {C} }
holomorphic function, then
z
0
{\displaystyle z_{0}}
isolated singularity of
f
{\displaystyle f}
Depending on the behavior of
f
{\displaystyle f}
z
0
{\displaystyle z_{0}}
f
{\displaystyle f}
If
f
{\displaystyle f}
G
{\displaystyle G}
z
0
{\displaystyle z_{0}}
removable singularity . According to the Riemann Removability Theorem , this is the case if
f
{\displaystyle f}
z
0
{\displaystyle z_{0}}
If
z
0
{\displaystyle z_{0}}
n
≥
1
{\displaystyle n\geq 1}
(
⋅
−
z
0
)
n
⋅
f
{\displaystyle (\cdot -z_{0})^{n}\cdot f}
z
0
{\displaystyle z_{0}}
f
{\displaystyle f}
pole at
z
0
{\displaystyle z_{0}}
n
{\displaystyle n}
order of the pole.
If
z
0
{\displaystyle z_{0}}
z
0
{\displaystyle z_{0}}
essential singularity of
f
{\displaystyle f}
Since
lim
z
→
0
sin
z
z
=
1
{\displaystyle \lim _{z\to 0}{\frac {\sin z}{z}}=1}
f
1
(
z
)
=
sin
z
z
{\displaystyle f_{1}(z)={\frac {\sin z}{z}}}
z
0
=
0
{\displaystyle z_{0}=0}
The function
f
2
(
z
)
=
1
sin
z
{\displaystyle f_{2}(z)={\frac {1}{\sin z}}}
z
0
=
0
{\displaystyle z_{0}=0}
f
2
{\displaystyle f_{2}}
0
{\displaystyle 0}
f
2
{\displaystyle f_{2}}
0
{\displaystyle 0}
f
2
(
z
)
⋅
(
z
−
0
)
1
=
f
2
(
z
)
z
=
z
sin
z
{\displaystyle f_{2}(z)\cdot (z-0)^{1}=f_{2}(z)z={\frac {z}{\sin z}}}
lim
z
→
0
z
sin
z
=
1
{\displaystyle \lim _{z\to 0}{\frac {z}{\sin z}}=1}
The function
f
3
(
z
)
=
sin
1
z
{\displaystyle f_{3}(z)=\sin {\frac {1}{z}}}
z
0
=
0
{\displaystyle z_{0}=0}
n
≥
1
{\displaystyle n\geq 1}
f
3
(
z
)
z
n
=
z
n
sin
1
z
{\displaystyle f_{3}(z)z^{n}=z^{n}\sin {\frac {1}{z}}}
0
{\displaystyle 0}
sin
z
−
1
=
e
i
z
−
1
−
e
−
i
z
−
1
2
i
{\displaystyle \sin z^{-1}={\frac {e^{iz^{-1}}-e^{-iz^{-1}}}{2i}}}
z
=
i
t
{\displaystyle z=it}
t
∈
R
{\displaystyle t\in \mathbb {R} }
f
3
(
i
t
)
(
i
t
)
n
=
(
i
t
)
n
e
t
−
1
−
e
−
t
−
1
2
i
{\displaystyle f_{3}(it)(it)^{n}=(it)^{n}{\frac {e^{t^{-1}}-e^{-t^{-1}}}{2i}}}
t
→
0
+
{\displaystyle t\to 0^{+}}
The type of isolated singularity can also be inferred from the Laurent Expansion of
f
{\displaystyle f}
z
0
{\displaystyle z_{0}}
f
(
z
)
=
∑
n
=
−
∞
∞
a
n
(
z
−
z
0
)
n
{\displaystyle f(z)=\sum _{n=-\infty }^{\infty }a_{n}(z-z_{0})^{n}}
be the Laurent Series of
f
{\displaystyle f}
z
0
{\displaystyle z_{0}}
o
z
(
f
)
=
sup
{
n
∈
Z
|
∀
k
<
n
:
a
k
=
0
}
{\displaystyle o_{z}(f)=\sup\{n\in \mathbb {Z} |\forall k<n:a_{k}=0\}}
Then,
f
{\displaystyle f}
o
z
(
f
)
≥
0
{\displaystyle o_{z}(f)\geq 0}
main part of the series is zero, and the singularity is removable.
−
∞
<
o
z
(
f
)
<
0
{\displaystyle -\infty <o_{z}(f)<0}
−
o
z
(
f
)
{\displaystyle -o_{z}(f)}
o
z
(
f
)
=
−
∞
{\displaystyle o_{z}(f)=-\infty }
Let us consider our three examples again:
It is
f
1
(
z
)
=
sin
z
z
=
∑
k
=
0
∞
(
−
1
)
n
z
2
n
(
2
n
+
1
)
!
{\displaystyle f_{1}(z)={\frac {\sin z}{z}}=\sum _{k=0}^{\infty }(-1)^{n}{\frac {z^{2n}}{(2n+1)!}}}
o
0
(
f
1
)
=
0
{\displaystyle o_{0}(f_{1})=0}
It is
f
2
(
z
)
=
1
sin
z
=
1
z
+
z
6
+
7
360
z
3
+
…
{\displaystyle f_{2}(z)={\frac {1}{\sin z}}={\frac {1}{z}}+{\frac {z}{6}}+{\frac {7}{360}}z^{3}+\ldots }
o
0
(
f
2
)
=
−
1
{\displaystyle o_{0}(f_{2})=-1}
It is
f
3
(
z
)
=
sin
z
−
1
=
∑
n
=
−
∞
0
(
−
1
)
n
(
−
2
n
+
1
)
!
z
2
n
−
1
{\displaystyle f_{3}(z)=\sin z^{-1}=\sum _{n=-\infty }^{0}{\frac {(-1)^{n}}{(-2n+1)!}}z^{2n-1}}
o
0
(
f
3
)
=
−
∞
{\displaystyle o_{0}(f_{3})=-\infty }
This page was translated based on the following Wikiversity source page and uses the concept of Translation and Version Control for a transparent language fork in a Wikiversity:
https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/isolierte_Singularität
