Can $frac{1}{z}$ be expressed as a Laurent series?
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Can the complex function $frac{1}{z}$ be expressed as a Laurent series, around $z=0$, over the whole complex plane? I have tried re-writing it as $frac{-1}{1-(z-1)}$. However I can only expand this under the condition that $left | z-1 right |geq 1$, which won't satisfy $left | z right |geq 0$.
It's probably worth noting that I'm actually trying to calculate the Laurent series of $frac{1}{z-i}$ around $z=i$ and was merely using $w=z-i$ as a substitution.
complex-analysis
asked Dec 3 '18 at 0:50
Relatively GeneralRelatively General
31
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add a comment |
$begingroup$
Can the complex function $frac{1}{z}$ be expressed as a Laurent series, around $z=0$, over the whole complex plane? I have tried re-writing it as $frac{-1}{1-(z-1)}$. However I can only expand this under the condition that $left | z-1 right |geq 1$, which won't satisfy $left | z right |geq 0$.
It's probably worth noting that I'm actually trying to calculate the Laurent series of $frac{1}{z-i}$ around $z=i$ and was merely using $w=z-i$ as a substitution.
complex-analysis
asked Dec 3 '18 at 0:50
Relatively GeneralRelatively General
31
$endgroup$
5
$begingroup$
$1/z= z^{-1}$ ...
$endgroup$
– Torsten Schoeneberg
Dec 3 '18 at 0:51
2
$begingroup$
it is already a Laurent series, it is its own...
$endgroup$
– user29418
Dec 3 '18 at 0:53
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$1/z$ already is a Laurent series (as @TorstenSchoeneberg comments).
$endgroup$
– Ethan Bolker
Dec 3 '18 at 0:54
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@Torsten Schoeneberg Oh. So would this imply that 1/(z-i) is also a Laurent series?
$endgroup$
– Relatively General
Dec 3 '18 at 0:57
4
$begingroup$
Well the Laurent series of $frac{1}{z-i}$ around $i$ is already $frac{1}{z-i}$.
$endgroup$
– Alonso Delfín
Dec 3 '18 at 0:57
add a comment |
$begingroup$
Can the complex function $frac{1}{z}$ be expressed as a Laurent series, around $z=0$, over the whole complex plane? I have tried re-writing it as $frac{-1}{1-(z-1)}$. However I can only expand this under the condition that $left | z-1 right |geq 1$, which won't satisfy $left | z right |geq 0$.
It's probably worth noting that I'm actually trying to calculate the Laurent series of $frac{1}{z-i}$ around $z=i$ and was merely using $w=z-i$ as a substitution.
complex-analysis
asked Dec 3 '18 at 0:50
Relatively GeneralRelatively General
31
$endgroup$
Can the complex function $frac{1}{z}$ be expressed as a Laurent series, around $z=0$, over the whole complex plane? I have tried re-writing it as $frac{-1}{1-(z-1)}$. However I can only expand this under the condition that $left | z-1 right |geq 1$, which won't satisfy $left | z right |geq 0$.
It's probably worth noting that I'm actually trying to calculate the Laurent series of $frac{1}{z-i}$ around $z=i$ and was merely using $w=z-i$ as a substitution.
complex-analysis
complex-analysis
asked Dec 3 '18 at 0:50
Relatively GeneralRelatively General
31
asked Dec 3 '18 at 0:50
Relatively GeneralRelatively General
31
asked Dec 3 '18 at 0:50
Relatively GeneralRelatively General
31
asked Dec 3 '18 at 0:50
Relatively GeneralRelatively General
31
asked Dec 3 '18 at 0:50
Relatively GeneralRelatively General
31
31
5
$begingroup$
$1/z= z^{-1}$ ...
$endgroup$
– Torsten Schoeneberg
Dec 3 '18 at 0:51
2
$begingroup$
it is already a Laurent series, it is its own...
$endgroup$
– user29418
Dec 3 '18 at 0:53
$begingroup$
$1/z$ already is a Laurent series (as @TorstenSchoeneberg comments).
$endgroup$
– Ethan Bolker
Dec 3 '18 at 0:54
$begingroup$
@Torsten Schoeneberg Oh. So would this imply that 1/(z-i) is also a Laurent series?
$endgroup$
– Relatively General
Dec 3 '18 at 0:57
4
$begingroup$
Well the Laurent series of $frac{1}{z-i}$ around $i$ is already $frac{1}{z-i}$.
$endgroup$
– Alonso Delfín
Dec 3 '18 at 0:57
add a comment |
5
$begingroup$
$1/z= z^{-1}$ ...
$endgroup$
– Torsten Schoeneberg
Dec 3 '18 at 0:51
2
$begingroup$
it is already a Laurent series, it is its own...
$endgroup$
– user29418
Dec 3 '18 at 0:53
$begingroup$
$1/z$ already is a Laurent series (as @TorstenSchoeneberg comments).
$endgroup$
– Ethan Bolker
Dec 3 '18 at 0:54
$begingroup$
@Torsten Schoeneberg Oh. So would this imply that 1/(z-i) is also a Laurent series?
$endgroup$
– Relatively General
Dec 3 '18 at 0:57
4
$begingroup$
Well the Laurent series of $frac{1}{z-i}$ around $i$ is already $frac{1}{z-i}$.
$endgroup$
– Alonso Delfín
Dec 3 '18 at 0:57
5
5
$begingroup$
$1/z= z^{-1}$ ...
$endgroup$
– Torsten Schoeneberg
Dec 3 '18 at 0:51
$begingroup$
$1/z= z^{-1}$ ...
$endgroup$
– Torsten Schoeneberg
Dec 3 '18 at 0:51
2
2
$begingroup$
it is already a Laurent series, it is its own...
$endgroup$
– user29418
Dec 3 '18 at 0:53
$begingroup$
it is already a Laurent series, it is its own...
$endgroup$
– user29418
Dec 3 '18 at 0:53
$begingroup$
$1/z$ already is a Laurent series (as @TorstenSchoeneberg comments).
$endgroup$
– Ethan Bolker
Dec 3 '18 at 0:54
$begingroup$
$1/z$ already is a Laurent series (as @TorstenSchoeneberg comments).
$endgroup$
– Ethan Bolker
Dec 3 '18 at 0:54
$begingroup$
@Torsten Schoeneberg Oh. So would this imply that 1/(z-i) is also a Laurent series?
$endgroup$
– Relatively General
Dec 3 '18 at 0:57
$begingroup$
@Torsten Schoeneberg Oh. So would this imply that 1/(z-i) is also a Laurent series?
$endgroup$
– Relatively General
Dec 3 '18 at 0:57
4
4
$begingroup$
Well the Laurent series of $frac{1}{z-i}$ around $i$ is already $frac{1}{z-i}$.
$endgroup$
– Alonso Delfín
Dec 3 '18 at 0:57
$begingroup$
Well the Laurent series of $frac{1}{z-i}$ around $i$ is already $frac{1}{z-i}$.
$endgroup$
– Alonso Delfín
Dec 3 '18 at 0:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is nothing to calculate in either case.
The Laurent series of $frac{1}{z-i}$ around $z=i$ is in the form:
begin{align}cdots+frac{a_{-n}}{(z-i)^n}+frac{a_{-n+1}}{(z-i)^{n-1}}+cdots+frac{a_{-1}}{z-i}+a_0+a_1(z-i)+a_2(z-i)^2+cdotsend{align}
with all coefficients equal to zero except $a_{-1}=1$.
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
There is nothing to calculate in either case.
The Laurent series of $frac{1}{z-i}$ around $z=i$ is in the form:
begin{align}cdots+frac{a_{-n}}{(z-i)^n}+frac{a_{-n+1}}{(z-i)^{n-1}}+cdots+frac{a_{-1}}{z-i}+a_0+a_1(z-i)+a_2(z-i)^2+cdotsend{align}
with all coefficients equal to zero except $a_{-1}=1$.
$endgroup$
add a comment |
$begingroup$
There is nothing to calculate in either case.
The Laurent series of $frac{1}{z-i}$ around $z=i$ is in the form:
begin{align}cdots+frac{a_{-n}}{(z-i)^n}+frac{a_{-n+1}}{(z-i)^{n-1}}+cdots+frac{a_{-1}}{z-i}+a_0+a_1(z-i)+a_2(z-i)^2+cdotsend{align}
with all coefficients equal to zero except $a_{-1}=1$.
$endgroup$
add a comment |
$begingroup$
There is nothing to calculate in either case.
The Laurent series of $frac{1}{z-i}$ around $z=i$ is in the form:
begin{align}cdots+frac{a_{-n}}{(z-i)^n}+frac{a_{-n+1}}{(z-i)^{n-1}}+cdots+frac{a_{-1}}{z-i}+a_0+a_1(z-i)+a_2(z-i)^2+cdotsend{align}
with all coefficients equal to zero except $a_{-1}=1$.
$endgroup$
There is nothing to calculate in either case.
The Laurent series of $frac{1}{z-i}$ around $z=i$ is in the form:
begin{align}cdots+frac{a_{-n}}{(z-i)^n}+frac{a_{-n+1}}{(z-i)^{n-1}}+cdots+frac{a_{-1}}{z-i}+a_0+a_1(z-i)+a_2(z-i)^2+cdotsend{align}
with all coefficients equal to zero except $a_{-1}=1$.
answered Dec 3 '18 at 11:24
user621367
add a comment |
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5
$begingroup$
$1/z= z^{-1}$ ...
$endgroup$
– Torsten Schoeneberg
Dec 3 '18 at 0:51
2
$begingroup$
it is already a Laurent series, it is its own...
$endgroup$
– user29418
Dec 3 '18 at 0:53
$begingroup$
$1/z$ already is a Laurent series (as @TorstenSchoeneberg comments).
$endgroup$
– Ethan Bolker
Dec 3 '18 at 0:54
$begingroup$
@Torsten Schoeneberg Oh. So would this imply that 1/(z-i) is also a Laurent series?
$endgroup$
– Relatively General
Dec 3 '18 at 0:57
4
$begingroup$
Well the Laurent series of $frac{1}{z-i}$ around $i$ is already $frac{1}{z-i}$.
$endgroup$
– Alonso Delfín
Dec 3 '18 at 0:57