Laurent Series expansion about the point $z_0 = i$ of $frac{z}{z^2+1}$
I am trying to construct the Laurent series expansion of $f(z) = frac{z}{z^2+1}$ about $z_0 = i$ in the region ${z in mathbb{C}: 0 < |z - i| < 2}$ but I am stuck.
We can re-write $f(z) = dfrac{z}{z^2+1} = dfrac{z}{(z-i)(z+i)}$
This allows us to see more easily that there are first order poles at $pm i$. We can use partial fractions to obtain:
$$frac{1}{(z+i)(z-i)} = frac{i}{2(z+i)} - frac{i}{2(z-i)}$$
We want to re-write everything in terms of $z-i$:
$$frac{1}{z+i} = frac{1}{2i + (z - i)} = frac{1}{2i}left(frac{1}{1+frac{1}{2i}(z-i)}right)$$
Using the geometric series this gives:
$$ = frac{-i}{2} left(sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right)$$
Then we have that:
$$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$
But the problem is, I still have a term in the form of $z$ rather than $z-i$. How do I proceed from here?
complex-analysis taylor-expansion laurent-series
edited Dec 2 '18 at 19:51
Lorenzo B.
1,8322520
asked Dec 2 '18 at 18:21
Jane Doe
18212
add a comment |
I am trying to construct the Laurent series expansion of $f(z) = frac{z}{z^2+1}$ about $z_0 = i$ in the region ${z in mathbb{C}: 0 < |z - i| < 2}$ but I am stuck.
We can re-write $f(z) = dfrac{z}{z^2+1} = dfrac{z}{(z-i)(z+i)}$
This allows us to see more easily that there are first order poles at $pm i$. We can use partial fractions to obtain:
$$frac{1}{(z+i)(z-i)} = frac{i}{2(z+i)} - frac{i}{2(z-i)}$$
We want to re-write everything in terms of $z-i$:
$$frac{1}{z+i} = frac{1}{2i + (z - i)} = frac{1}{2i}left(frac{1}{1+frac{1}{2i}(z-i)}right)$$
Using the geometric series this gives:
$$ = frac{-i}{2} left(sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right)$$
Then we have that:
$$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$
But the problem is, I still have a term in the form of $z$ rather than $z-i$. How do I proceed from here?
complex-analysis taylor-expansion laurent-series
edited Dec 2 '18 at 19:51
Lorenzo B.
1,8322520
asked Dec 2 '18 at 18:21
Jane Doe
18212
$f(i)$ is not defined
– Tito Eliatron
Dec 2 '18 at 18:27
1
@TitoEliatron You’re correct but that’s the use of Laurent series howellkb.uah.edu/MathPhysicsText/Complex_Variables/Laurent.pdf
– Jane Doe
Dec 2 '18 at 18:33
Sorry, I read Taylor, not Laurent.
– Tito Eliatron
Dec 2 '18 at 18:38
You can write $z=(z-i)+i$ and proceed
– Tito Eliatron
Dec 2 '18 at 18:39
add a comment |
I am trying to construct the Laurent series expansion of $f(z) = frac{z}{z^2+1}$ about $z_0 = i$ in the region ${z in mathbb{C}: 0 < |z - i| < 2}$ but I am stuck.
We can re-write $f(z) = dfrac{z}{z^2+1} = dfrac{z}{(z-i)(z+i)}$
This allows us to see more easily that there are first order poles at $pm i$. We can use partial fractions to obtain:
$$frac{1}{(z+i)(z-i)} = frac{i}{2(z+i)} - frac{i}{2(z-i)}$$
We want to re-write everything in terms of $z-i$:
$$frac{1}{z+i} = frac{1}{2i + (z - i)} = frac{1}{2i}left(frac{1}{1+frac{1}{2i}(z-i)}right)$$
Using the geometric series this gives:
$$ = frac{-i}{2} left(sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right)$$
Then we have that:
$$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$
But the problem is, I still have a term in the form of $z$ rather than $z-i$. How do I proceed from here?
complex-analysis taylor-expansion laurent-series
edited Dec 2 '18 at 19:51
Lorenzo B.
1,8322520
asked Dec 2 '18 at 18:21
Jane Doe
18212
I am trying to construct the Laurent series expansion of $f(z) = frac{z}{z^2+1}$ about $z_0 = i$ in the region ${z in mathbb{C}: 0 < |z - i| < 2}$ but I am stuck.
We can re-write $f(z) = dfrac{z}{z^2+1} = dfrac{z}{(z-i)(z+i)}$
This allows us to see more easily that there are first order poles at $pm i$. We can use partial fractions to obtain:
$$frac{1}{(z+i)(z-i)} = frac{i}{2(z+i)} - frac{i}{2(z-i)}$$
We want to re-write everything in terms of $z-i$:
$$frac{1}{z+i} = frac{1}{2i + (z - i)} = frac{1}{2i}left(frac{1}{1+frac{1}{2i}(z-i)}right)$$
Using the geometric series this gives:
$$ = frac{-i}{2} left(sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right)$$
Then we have that:
$$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$
But the problem is, I still have a term in the form of $z$ rather than $z-i$. How do I proceed from here?
complex-analysis taylor-expansion laurent-series
complex-analysis taylor-expansion laurent-series
edited Dec 2 '18 at 19:51
Lorenzo B.
1,8322520
asked Dec 2 '18 at 18:21
Jane Doe
18212
edited Dec 2 '18 at 19:51
Lorenzo B.
1,8322520
asked Dec 2 '18 at 18:21
Jane Doe
18212
edited Dec 2 '18 at 19:51
Lorenzo B.
1,8322520
edited Dec 2 '18 at 19:51
Lorenzo B.
1,8322520
edited Dec 2 '18 at 19:51
Lorenzo B.
1,8322520
1,8322520
asked Dec 2 '18 at 18:21
Jane Doe
18212
asked Dec 2 '18 at 18:21
Jane Doe
18212
asked Dec 2 '18 at 18:21
Jane Doe
18212
18212
$f(i)$ is not defined
– Tito Eliatron
Dec 2 '18 at 18:27
1
@TitoEliatron You’re correct but that’s the use of Laurent series howellkb.uah.edu/MathPhysicsText/Complex_Variables/Laurent.pdf
– Jane Doe
Dec 2 '18 at 18:33
Sorry, I read Taylor, not Laurent.
– Tito Eliatron
Dec 2 '18 at 18:38
You can write $z=(z-i)+i$ and proceed
– Tito Eliatron
Dec 2 '18 at 18:39
add a comment |
$f(i)$ is not defined
– Tito Eliatron
Dec 2 '18 at 18:27
1
@TitoEliatron You’re correct but that’s the use of Laurent series howellkb.uah.edu/MathPhysicsText/Complex_Variables/Laurent.pdf
– Jane Doe
Dec 2 '18 at 18:33
Sorry, I read Taylor, not Laurent.
– Tito Eliatron
Dec 2 '18 at 18:38
You can write $z=(z-i)+i$ and proceed
– Tito Eliatron
Dec 2 '18 at 18:39
$f(i)$ is not defined
– Tito Eliatron
Dec 2 '18 at 18:27
$f(i)$ is not defined
– Tito Eliatron
Dec 2 '18 at 18:27
1
1
@TitoEliatron You’re correct but that’s the use of Laurent series howellkb.uah.edu/MathPhysicsText/Complex_Variables/Laurent.pdf
– Jane Doe
Dec 2 '18 at 18:33
@TitoEliatron You’re correct but that’s the use of Laurent series howellkb.uah.edu/MathPhysicsText/Complex_Variables/Laurent.pdf
– Jane Doe
Dec 2 '18 at 18:33
Sorry, I read Taylor, not Laurent.
– Tito Eliatron
Dec 2 '18 at 18:38
Sorry, I read Taylor, not Laurent.
– Tito Eliatron
Dec 2 '18 at 18:38
You can write $z=(z-i)+i$ and proceed
– Tito Eliatron
Dec 2 '18 at 18:39
You can write $z=(z-i)+i$ and proceed
– Tito Eliatron
Dec 2 '18 at 18:39
add a comment |
2 Answers
2
active
oldest
votes
begin{align}&f(z)=frac{z}{z^2+1}=frac{A}{z-i}+frac{B}{z+i}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{z+i}end{align}
begin{align}&A=Res[f,i]=limlimits_{zrightarrow i}frac{z}{z+i}=frac{1}{2}
\&B=Res[f,-i]=limlimits_{zrightarrow -i}frac{z}{z-i}=frac{1}{2}end{align}begin{align}f(z)&= frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{2ileft(1+frac{z-i}{2i}right)}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}sum_{n=0}^{infty}frac{(-1)^n (z-i)^n}{(2i)^{n+1}}end{align}begin{align}=&frac{1}{2}frac{1}{z-i}-frac{i}{4}+frac{1}{8}(z-i)\&+frac{i}{16}(z-i)^2+cdotsend{align}
in the region ${zinmathbb{C}:0<|z-i|<2}.$
Thank you, that way makes much more sense!
– Jane Doe
Dec 3 '18 at 15:27
add a comment |
Following your argument,
$$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right) = (z-i+i)left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)\ = left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^{n+1}}{(2i)^n}right) - 1right) -left(frac{1}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$
answered Dec 2 '18 at 18:42
Tito Eliatron
1,450622
add a comment |
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2 Answers
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2 Answers
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begin{align}&f(z)=frac{z}{z^2+1}=frac{A}{z-i}+frac{B}{z+i}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{z+i}end{align}
begin{align}&A=Res[f,i]=limlimits_{zrightarrow i}frac{z}{z+i}=frac{1}{2}
\&B=Res[f,-i]=limlimits_{zrightarrow -i}frac{z}{z-i}=frac{1}{2}end{align}begin{align}f(z)&= frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{2ileft(1+frac{z-i}{2i}right)}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}sum_{n=0}^{infty}frac{(-1)^n (z-i)^n}{(2i)^{n+1}}end{align}begin{align}=&frac{1}{2}frac{1}{z-i}-frac{i}{4}+frac{1}{8}(z-i)\&+frac{i}{16}(z-i)^2+cdotsend{align}
in the region ${zinmathbb{C}:0<|z-i|<2}.$
Thank you, that way makes much more sense!
– Jane Doe
Dec 3 '18 at 15:27
add a comment |
begin{align}&f(z)=frac{z}{z^2+1}=frac{A}{z-i}+frac{B}{z+i}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{z+i}end{align}
begin{align}&A=Res[f,i]=limlimits_{zrightarrow i}frac{z}{z+i}=frac{1}{2}
\&B=Res[f,-i]=limlimits_{zrightarrow -i}frac{z}{z-i}=frac{1}{2}end{align}begin{align}f(z)&= frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{2ileft(1+frac{z-i}{2i}right)}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}sum_{n=0}^{infty}frac{(-1)^n (z-i)^n}{(2i)^{n+1}}end{align}begin{align}=&frac{1}{2}frac{1}{z-i}-frac{i}{4}+frac{1}{8}(z-i)\&+frac{i}{16}(z-i)^2+cdotsend{align}
in the region ${zinmathbb{C}:0<|z-i|<2}.$
Thank you, that way makes much more sense!
– Jane Doe
Dec 3 '18 at 15:27
add a comment |
begin{align}&f(z)=frac{z}{z^2+1}=frac{A}{z-i}+frac{B}{z+i}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{z+i}end{align}
begin{align}&A=Res[f,i]=limlimits_{zrightarrow i}frac{z}{z+i}=frac{1}{2}
\&B=Res[f,-i]=limlimits_{zrightarrow -i}frac{z}{z-i}=frac{1}{2}end{align}begin{align}f(z)&= frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{2ileft(1+frac{z-i}{2i}right)}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}sum_{n=0}^{infty}frac{(-1)^n (z-i)^n}{(2i)^{n+1}}end{align}begin{align}=&frac{1}{2}frac{1}{z-i}-frac{i}{4}+frac{1}{8}(z-i)\&+frac{i}{16}(z-i)^2+cdotsend{align}
in the region ${zinmathbb{C}:0<|z-i|<2}.$
begin{align}&f(z)=frac{z}{z^2+1}=frac{A}{z-i}+frac{B}{z+i}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{z+i}end{align}
begin{align}&A=Res[f,i]=limlimits_{zrightarrow i}frac{z}{z+i}=frac{1}{2}
\&B=Res[f,-i]=limlimits_{zrightarrow -i}frac{z}{z-i}=frac{1}{2}end{align}begin{align}f(z)&= frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{2ileft(1+frac{z-i}{2i}right)}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}sum_{n=0}^{infty}frac{(-1)^n (z-i)^n}{(2i)^{n+1}}end{align}begin{align}=&frac{1}{2}frac{1}{z-i}-frac{i}{4}+frac{1}{8}(z-i)\&+frac{i}{16}(z-i)^2+cdotsend{align}
in the region ${zinmathbb{C}:0<|z-i|<2}.$
answered Dec 2 '18 at 22:19
user621367
Thank you, that way makes much more sense!
– Jane Doe
Dec 3 '18 at 15:27
add a comment |
Thank you, that way makes much more sense!
– Jane Doe
Dec 3 '18 at 15:27
Thank you, that way makes much more sense!
– Jane Doe
Dec 3 '18 at 15:27
Thank you, that way makes much more sense!
– Jane Doe
Dec 3 '18 at 15:27
add a comment |
Following your argument,
$$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right) = (z-i+i)left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)\ = left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^{n+1}}{(2i)^n}right) - 1right) -left(frac{1}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$
answered Dec 2 '18 at 18:42
Tito Eliatron
1,450622
add a comment |
Following your argument,
$$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right) = (z-i+i)left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)\ = left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^{n+1}}{(2i)^n}right) - 1right) -left(frac{1}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$
answered Dec 2 '18 at 18:42
Tito Eliatron
1,450622
add a comment |
Following your argument,
$$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right) = (z-i+i)left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)\ = left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^{n+1}}{(2i)^n}right) - 1right) -left(frac{1}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$
answered Dec 2 '18 at 18:42
Tito Eliatron
1,450622
Following your argument,
$$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right) = (z-i+i)left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)\ = left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^{n+1}}{(2i)^n}right) - 1right) -left(frac{1}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$
answered Dec 2 '18 at 18:42
Tito Eliatron
1,450622
answered Dec 2 '18 at 18:42
Tito Eliatron
1,450622
answered Dec 2 '18 at 18:42
Tito Eliatron
1,450622
answered Dec 2 '18 at 18:42
Tito Eliatron
1,450622
1,450622
add a comment |
add a comment |
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$f(i)$ is not defined
– Tito Eliatron
Dec 2 '18 at 18:27
1
@TitoEliatron You’re correct but that’s the use of Laurent series howellkb.uah.edu/MathPhysicsText/Complex_Variables/Laurent.pdf
– Jane Doe
Dec 2 '18 at 18:33
Sorry, I read Taylor, not Laurent.
– Tito Eliatron
Dec 2 '18 at 18:38
You can write $z=(z-i)+i$ and proceed
– Tito Eliatron
Dec 2 '18 at 18:39