General strategy in finding Laurent Series
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I'm taking my first course in complex analysis and I understand everything but how to find Laurent series. Other than seeing worked examples I can never figure them out myself.
One of the textbook problems I'm stuck on is:
Find the Laurent series expansion of:
$$ frac{1}{(z^2 - 1)^2} $$
Defined at $|z+1|> 2$
What is the general strategy for approaching this problem? Any hints, guidance or resources would be greatly appreciated.
complex-analysis laurent-series
edited Nov 15 at 16:31
José Carlos Santos
140k18111203
asked Nov 15 at 16:22
Safder
1239
add a comment |
up vote
0
down vote
favorite
I'm taking my first course in complex analysis and I understand everything but how to find Laurent series. Other than seeing worked examples I can never figure them out myself.
One of the textbook problems I'm stuck on is:
Find the Laurent series expansion of:
$$ frac{1}{(z^2 - 1)^2} $$
Defined at $|z+1|> 2$
What is the general strategy for approaching this problem? Any hints, guidance or resources would be greatly appreciated.
complex-analysis laurent-series
edited Nov 15 at 16:31
José Carlos Santos
140k18111203
asked Nov 15 at 16:22
Safder
1239
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm taking my first course in complex analysis and I understand everything but how to find Laurent series. Other than seeing worked examples I can never figure them out myself.
One of the textbook problems I'm stuck on is:
Find the Laurent series expansion of:
$$ frac{1}{(z^2 - 1)^2} $$
Defined at $|z+1|> 2$
What is the general strategy for approaching this problem? Any hints, guidance or resources would be greatly appreciated.
complex-analysis laurent-series
edited Nov 15 at 16:31
José Carlos Santos
140k18111203
asked Nov 15 at 16:22
Safder
1239
I'm taking my first course in complex analysis and I understand everything but how to find Laurent series. Other than seeing worked examples I can never figure them out myself.
One of the textbook problems I'm stuck on is:
Find the Laurent series expansion of:
$$ frac{1}{(z^2 - 1)^2} $$
Defined at $|z+1|> 2$
What is the general strategy for approaching this problem? Any hints, guidance or resources would be greatly appreciated.
complex-analysis laurent-series
complex-analysis laurent-series
edited Nov 15 at 16:31
José Carlos Santos
140k18111203
asked Nov 15 at 16:22
Safder
1239
edited Nov 15 at 16:31
José Carlos Santos
140k18111203
asked Nov 15 at 16:22
Safder
1239
edited Nov 15 at 16:31
José Carlos Santos
140k18111203
edited Nov 15 at 16:31
José Carlos Santos
140k18111203
edited Nov 15 at 16:31
José Carlos Santos
140k18111203
140k18111203
asked Nov 15 at 16:22
Safder
1239
asked Nov 15 at 16:22
Safder
1239
asked Nov 15 at 16:22
Safder
1239
1239
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Express your function in terms fo $z+1$. In this case, you have$$frac1{(z^2-1)^2}=frac1{(z-1)^2(z+1)^2}=frac1{bigl(2-(z+1)bigr)^2(z+1)^2}.$$Now, at this point don't deal with the $dfrac1{(z+1)^2}$ parte; leave it to the end. Write $dfrac1{bigl(2-(z+1)bigr)^2}$ as a power series about $-1$:$$frac1{bigl(2-(z+1)bigr)^2}=a_0+a_1(z+1)+a_2(z+1)^2+a_3(z+1)^3+cdots$$Then the Laurent series that you are interested in is$$frac{a_0}{(z+1)^2}+frac{a_1}{z+1}+a_2+a_3(z+1)+cdots$$
answered Nov 15 at 16:28
José Carlos Santos
140k18111203
Okay I see what you are doing. Then you just multiplied in the other $frac{1}{(z+1)^2}$?
– Safder
Nov 15 at 16:36
Yes. That was my last step.
– José Carlos Santos
Nov 15 at 16:47
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Express your function in terms fo $z+1$. In this case, you have$$frac1{(z^2-1)^2}=frac1{(z-1)^2(z+1)^2}=frac1{bigl(2-(z+1)bigr)^2(z+1)^2}.$$Now, at this point don't deal with the $dfrac1{(z+1)^2}$ parte; leave it to the end. Write $dfrac1{bigl(2-(z+1)bigr)^2}$ as a power series about $-1$:$$frac1{bigl(2-(z+1)bigr)^2}=a_0+a_1(z+1)+a_2(z+1)^2+a_3(z+1)^3+cdots$$Then the Laurent series that you are interested in is$$frac{a_0}{(z+1)^2}+frac{a_1}{z+1}+a_2+a_3(z+1)+cdots$$
answered Nov 15 at 16:28
José Carlos Santos
140k18111203
Okay I see what you are doing. Then you just multiplied in the other $frac{1}{(z+1)^2}$?
– Safder
Nov 15 at 16:36
Yes. That was my last step.
– José Carlos Santos
Nov 15 at 16:47
add a comment |
up vote
1
down vote
accepted
Express your function in terms fo $z+1$. In this case, you have$$frac1{(z^2-1)^2}=frac1{(z-1)^2(z+1)^2}=frac1{bigl(2-(z+1)bigr)^2(z+1)^2}.$$Now, at this point don't deal with the $dfrac1{(z+1)^2}$ parte; leave it to the end. Write $dfrac1{bigl(2-(z+1)bigr)^2}$ as a power series about $-1$:$$frac1{bigl(2-(z+1)bigr)^2}=a_0+a_1(z+1)+a_2(z+1)^2+a_3(z+1)^3+cdots$$Then the Laurent series that you are interested in is$$frac{a_0}{(z+1)^2}+frac{a_1}{z+1}+a_2+a_3(z+1)+cdots$$
answered Nov 15 at 16:28
José Carlos Santos
140k18111203
Okay I see what you are doing. Then you just multiplied in the other $frac{1}{(z+1)^2}$?
– Safder
Nov 15 at 16:36
Yes. That was my last step.
– José Carlos Santos
Nov 15 at 16:47
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Express your function in terms fo $z+1$. In this case, you have$$frac1{(z^2-1)^2}=frac1{(z-1)^2(z+1)^2}=frac1{bigl(2-(z+1)bigr)^2(z+1)^2}.$$Now, at this point don't deal with the $dfrac1{(z+1)^2}$ parte; leave it to the end. Write $dfrac1{bigl(2-(z+1)bigr)^2}$ as a power series about $-1$:$$frac1{bigl(2-(z+1)bigr)^2}=a_0+a_1(z+1)+a_2(z+1)^2+a_3(z+1)^3+cdots$$Then the Laurent series that you are interested in is$$frac{a_0}{(z+1)^2}+frac{a_1}{z+1}+a_2+a_3(z+1)+cdots$$
answered Nov 15 at 16:28
José Carlos Santos
140k18111203
Express your function in terms fo $z+1$. In this case, you have$$frac1{(z^2-1)^2}=frac1{(z-1)^2(z+1)^2}=frac1{bigl(2-(z+1)bigr)^2(z+1)^2}.$$Now, at this point don't deal with the $dfrac1{(z+1)^2}$ parte; leave it to the end. Write $dfrac1{bigl(2-(z+1)bigr)^2}$ as a power series about $-1$:$$frac1{bigl(2-(z+1)bigr)^2}=a_0+a_1(z+1)+a_2(z+1)^2+a_3(z+1)^3+cdots$$Then the Laurent series that you are interested in is$$frac{a_0}{(z+1)^2}+frac{a_1}{z+1}+a_2+a_3(z+1)+cdots$$
answered Nov 15 at 16:28
José Carlos Santos
140k18111203
answered Nov 15 at 16:28
José Carlos Santos
140k18111203
answered Nov 15 at 16:28
José Carlos Santos
140k18111203
answered Nov 15 at 16:28
José Carlos Santos
140k18111203
140k18111203
Okay I see what you are doing. Then you just multiplied in the other $frac{1}{(z+1)^2}$?
– Safder
Nov 15 at 16:36
Yes. That was my last step.
– José Carlos Santos
Nov 15 at 16:47
add a comment |
Okay I see what you are doing. Then you just multiplied in the other $frac{1}{(z+1)^2}$?
– Safder
Nov 15 at 16:36
Yes. That was my last step.
– José Carlos Santos
Nov 15 at 16:47
Okay I see what you are doing. Then you just multiplied in the other $frac{1}{(z+1)^2}$?
– Safder
Nov 15 at 16:36
Okay I see what you are doing. Then you just multiplied in the other $frac{1}{(z+1)^2}$?
– Safder
Nov 15 at 16:36
Yes. That was my last step.
– José Carlos Santos
Nov 15 at 16:47
Yes. That was my last step.
– José Carlos Santos
Nov 15 at 16:47
add a comment |
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