Finding the Laurent series (complex numbers)
I have
$$
f(z)={frac{1}{z(1-z)}}
$$
Need to find the Laurent series around $z=0, z=1, z=infty$.
I did
$$
{frac{1}{z(1-z)}} = {frac{A}{z}}+{frac{B}{1-z}}
$$
and found $A=1, B=1$. Therefore we get
$$
{frac{1}{z}}+{frac{1}{1-z}} = {frac{1}{z}} + sum z^n
$$
But in the book this is the answer only for $z=0$. How should I find the answers for the other two? Thanks.
complex-analysis complex-numbers laurent-series
asked Dec 18 '18 at 17:56
user3132457
1336
add a comment |
I have
$$
f(z)={frac{1}{z(1-z)}}
$$
Need to find the Laurent series around $z=0, z=1, z=infty$.
I did
$$
{frac{1}{z(1-z)}} = {frac{A}{z}}+{frac{B}{1-z}}
$$
and found $A=1, B=1$. Therefore we get
$$
{frac{1}{z}}+{frac{1}{1-z}} = {frac{1}{z}} + sum z^n
$$
But in the book this is the answer only for $z=0$. How should I find the answers for the other two? Thanks.
complex-analysis complex-numbers laurent-series
asked Dec 18 '18 at 17:56
user3132457
1336
add a comment |
I have
$$
f(z)={frac{1}{z(1-z)}}
$$
Need to find the Laurent series around $z=0, z=1, z=infty$.
I did
$$
{frac{1}{z(1-z)}} = {frac{A}{z}}+{frac{B}{1-z}}
$$
and found $A=1, B=1$. Therefore we get
$$
{frac{1}{z}}+{frac{1}{1-z}} = {frac{1}{z}} + sum z^n
$$
But in the book this is the answer only for $z=0$. How should I find the answers for the other two? Thanks.
complex-analysis complex-numbers laurent-series
asked Dec 18 '18 at 17:56
user3132457
1336
I have
$$
f(z)={frac{1}{z(1-z)}}
$$
Need to find the Laurent series around $z=0, z=1, z=infty$.
I did
$$
{frac{1}{z(1-z)}} = {frac{A}{z}}+{frac{B}{1-z}}
$$
and found $A=1, B=1$. Therefore we get
$$
{frac{1}{z}}+{frac{1}{1-z}} = {frac{1}{z}} + sum z^n
$$
But in the book this is the answer only for $z=0$. How should I find the answers for the other two? Thanks.
complex-analysis complex-numbers laurent-series
complex-analysis complex-numbers laurent-series
asked Dec 18 '18 at 17:56
user3132457
1336
asked Dec 18 '18 at 17:56
user3132457
1336
asked Dec 18 '18 at 17:56
user3132457
1336
asked Dec 18 '18 at 17:56
user3132457
1336
asked Dec 18 '18 at 17:56
user3132457
1336
1336
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Hints :
For the $z=1$ case :
You need to create terms of the form $z-1$. You can manipulate your fraction decomposition that you already carried out, as :
$$f(z) = frac{1}{z(1-z)} = frac{1}{z} + frac{1}{1-z} = frac{1}{1+(z-1)} + frac{1}{1-z} $$
$$=$$
$$frac{1}{(z-1)left(frac{1}{z-1} + 1right)} - frac{1}{z-1} = frac{1}{z-1}left(frac{1}{frac{1}{z-1} + 1}right)$$
Now, recall the geometric series $frac{1}{1+w} = sum_{n=1}^infty (-1)^nw^n$. Let $w = frac{1}{z-1}$. Thus :
$$f(z) = frac{1}{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n =sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}$$
For the $infty$ case :
Recall the geometric series $frac{1}{1-w} = sum_{n=1}^infty w^{n}$ when $|w| <1$. Thus, for $|z| > 1$, we can write :
$$f(z) = frac{1}{z(1-z)}= -frac{1}{z^2(1-frac{1}{z})}=-sum_{n=0}^{infty}z^{-n-2}$$
Alternative : Let $w = 1/z$ and calculate the Laurent Series for $w =0$ which happens when $z to infty$.
answered Dec 18 '18 at 18:02
Rebellos
14.4k31245
That is not the LS around $z=1$. You have the Taylor series around $z=0$, for $f$ for $|z|<1$. See my posted solution for the LS around $z=1$.
– Mark Viola
Dec 18 '18 at 18:13
@MarkViola I am working in an edit from earlier since I saw my error.
– Rebellos
Dec 18 '18 at 18:14
@MarkViola I was editing before I even saw your solution or comment and rushed to assure you that I was already correcting my answer. I was operating via cellphone initially.
– Rebellos
Dec 18 '18 at 18:21
I understand. I've posted on MSE using a "not-so-smart-phone" many a time and it is definitely a challenge.
– Mark Viola
Dec 18 '18 at 18:24
add a comment |
In the annulus $1<|z|<infty$, we have
$$begin{align}
frac{1}{z(1-z)}&=frac{1}{z}+frac1{1-z}\\
&=frac{1}{1+(z-1)}+frac1{1-z}\\
&=frac1{z-1}frac{1}{1+frac1{z-1}}-frac1{z-1}\\
&=frac1{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n-frac1{z-1}\\
&=sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}
end{align}$$
answered Dec 18 '18 at 18:10
Mark Viola
130k1274170
add a comment |
We have
$$
eqalign{
& {1 over {zleft( {1 - z} right)}} = cr
& = left{ matrix{
- left( {{1 over z} + {1 over {left( {1 - z} right)}}} right)quad
Rightarrow quad - {1 over z} - sumlimits_{0, le ,n} {z^{,n} } quad left| {,z to 0} right. hfill cr
{1 over {left( {z - 1} right)}} - {1 over {left( {1 + left( {z - 1} right)} right)}}quad Rightarrow quad {1 over {left( {z - 1} right)}}
- sumlimits_{0, le ,n} {left( { - 1} right)^{,n} left( {z - 1} right)^{,n} } quad left| {,z to 1} right. hfill cr
- left( {{1 over z}} right)left( {1 - {1 over {left( {1 - {1 over z}} right)}}} right)quad
Rightarrow quad sumlimits_{0, le ,n} {left( {{1 over z}} right)^{,n + 2} } quad left| {,z to infty } right. hfill cr} right. cr}
$$
answered Dec 18 '18 at 18:28
G Cab
17.9k31237
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hints :
For the $z=1$ case :
You need to create terms of the form $z-1$. You can manipulate your fraction decomposition that you already carried out, as :
$$f(z) = frac{1}{z(1-z)} = frac{1}{z} + frac{1}{1-z} = frac{1}{1+(z-1)} + frac{1}{1-z} $$
$$=$$
$$frac{1}{(z-1)left(frac{1}{z-1} + 1right)} - frac{1}{z-1} = frac{1}{z-1}left(frac{1}{frac{1}{z-1} + 1}right)$$
Now, recall the geometric series $frac{1}{1+w} = sum_{n=1}^infty (-1)^nw^n$. Let $w = frac{1}{z-1}$. Thus :
$$f(z) = frac{1}{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n =sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}$$
For the $infty$ case :
Recall the geometric series $frac{1}{1-w} = sum_{n=1}^infty w^{n}$ when $|w| <1$. Thus, for $|z| > 1$, we can write :
$$f(z) = frac{1}{z(1-z)}= -frac{1}{z^2(1-frac{1}{z})}=-sum_{n=0}^{infty}z^{-n-2}$$
Alternative : Let $w = 1/z$ and calculate the Laurent Series for $w =0$ which happens when $z to infty$.
answered Dec 18 '18 at 18:02
Rebellos
14.4k31245
That is not the LS around $z=1$. You have the Taylor series around $z=0$, for $f$ for $|z|<1$. See my posted solution for the LS around $z=1$.
– Mark Viola
Dec 18 '18 at 18:13
@MarkViola I am working in an edit from earlier since I saw my error.
– Rebellos
Dec 18 '18 at 18:14
@MarkViola I was editing before I even saw your solution or comment and rushed to assure you that I was already correcting my answer. I was operating via cellphone initially.
– Rebellos
Dec 18 '18 at 18:21
I understand. I've posted on MSE using a "not-so-smart-phone" many a time and it is definitely a challenge.
– Mark Viola
Dec 18 '18 at 18:24
add a comment |
Hints :
For the $z=1$ case :
You need to create terms of the form $z-1$. You can manipulate your fraction decomposition that you already carried out, as :
$$f(z) = frac{1}{z(1-z)} = frac{1}{z} + frac{1}{1-z} = frac{1}{1+(z-1)} + frac{1}{1-z} $$
$$=$$
$$frac{1}{(z-1)left(frac{1}{z-1} + 1right)} - frac{1}{z-1} = frac{1}{z-1}left(frac{1}{frac{1}{z-1} + 1}right)$$
Now, recall the geometric series $frac{1}{1+w} = sum_{n=1}^infty (-1)^nw^n$. Let $w = frac{1}{z-1}$. Thus :
$$f(z) = frac{1}{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n =sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}$$
For the $infty$ case :
Recall the geometric series $frac{1}{1-w} = sum_{n=1}^infty w^{n}$ when $|w| <1$. Thus, for $|z| > 1$, we can write :
$$f(z) = frac{1}{z(1-z)}= -frac{1}{z^2(1-frac{1}{z})}=-sum_{n=0}^{infty}z^{-n-2}$$
Alternative : Let $w = 1/z$ and calculate the Laurent Series for $w =0$ which happens when $z to infty$.
answered Dec 18 '18 at 18:02
Rebellos
14.4k31245
That is not the LS around $z=1$. You have the Taylor series around $z=0$, for $f$ for $|z|<1$. See my posted solution for the LS around $z=1$.
– Mark Viola
Dec 18 '18 at 18:13
@MarkViola I am working in an edit from earlier since I saw my error.
– Rebellos
Dec 18 '18 at 18:14
@MarkViola I was editing before I even saw your solution or comment and rushed to assure you that I was already correcting my answer. I was operating via cellphone initially.
– Rebellos
Dec 18 '18 at 18:21
I understand. I've posted on MSE using a "not-so-smart-phone" many a time and it is definitely a challenge.
– Mark Viola
Dec 18 '18 at 18:24
add a comment |
Hints :
For the $z=1$ case :
You need to create terms of the form $z-1$. You can manipulate your fraction decomposition that you already carried out, as :
$$f(z) = frac{1}{z(1-z)} = frac{1}{z} + frac{1}{1-z} = frac{1}{1+(z-1)} + frac{1}{1-z} $$
$$=$$
$$frac{1}{(z-1)left(frac{1}{z-1} + 1right)} - frac{1}{z-1} = frac{1}{z-1}left(frac{1}{frac{1}{z-1} + 1}right)$$
Now, recall the geometric series $frac{1}{1+w} = sum_{n=1}^infty (-1)^nw^n$. Let $w = frac{1}{z-1}$. Thus :
$$f(z) = frac{1}{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n =sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}$$
For the $infty$ case :
Recall the geometric series $frac{1}{1-w} = sum_{n=1}^infty w^{n}$ when $|w| <1$. Thus, for $|z| > 1$, we can write :
$$f(z) = frac{1}{z(1-z)}= -frac{1}{z^2(1-frac{1}{z})}=-sum_{n=0}^{infty}z^{-n-2}$$
Alternative : Let $w = 1/z$ and calculate the Laurent Series for $w =0$ which happens when $z to infty$.
answered Dec 18 '18 at 18:02
Rebellos
14.4k31245
Hints :
For the $z=1$ case :
You need to create terms of the form $z-1$. You can manipulate your fraction decomposition that you already carried out, as :
$$f(z) = frac{1}{z(1-z)} = frac{1}{z} + frac{1}{1-z} = frac{1}{1+(z-1)} + frac{1}{1-z} $$
$$=$$
$$frac{1}{(z-1)left(frac{1}{z-1} + 1right)} - frac{1}{z-1} = frac{1}{z-1}left(frac{1}{frac{1}{z-1} + 1}right)$$
Now, recall the geometric series $frac{1}{1+w} = sum_{n=1}^infty (-1)^nw^n$. Let $w = frac{1}{z-1}$. Thus :
$$f(z) = frac{1}{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n =sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}$$
For the $infty$ case :
Recall the geometric series $frac{1}{1-w} = sum_{n=1}^infty w^{n}$ when $|w| <1$. Thus, for $|z| > 1$, we can write :
$$f(z) = frac{1}{z(1-z)}= -frac{1}{z^2(1-frac{1}{z})}=-sum_{n=0}^{infty}z^{-n-2}$$
Alternative : Let $w = 1/z$ and calculate the Laurent Series for $w =0$ which happens when $z to infty$.
answered Dec 18 '18 at 18:02
Rebellos
14.4k31245
edited Dec 18 '18 at 18:18
answered Dec 18 '18 at 18:02
Rebellos
14.4k31245
answered Dec 18 '18 at 18:02
Rebellos
14.4k31245
answered Dec 18 '18 at 18:02
Rebellos
14.4k31245
14.4k31245
That is not the LS around $z=1$. You have the Taylor series around $z=0$, for $f$ for $|z|<1$. See my posted solution for the LS around $z=1$.
– Mark Viola
Dec 18 '18 at 18:13
@MarkViola I am working in an edit from earlier since I saw my error.
– Rebellos
Dec 18 '18 at 18:14
@MarkViola I was editing before I even saw your solution or comment and rushed to assure you that I was already correcting my answer. I was operating via cellphone initially.
– Rebellos
Dec 18 '18 at 18:21
I understand. I've posted on MSE using a "not-so-smart-phone" many a time and it is definitely a challenge.
– Mark Viola
Dec 18 '18 at 18:24
add a comment |
That is not the LS around $z=1$. You have the Taylor series around $z=0$, for $f$ for $|z|<1$. See my posted solution for the LS around $z=1$.
– Mark Viola
Dec 18 '18 at 18:13
@MarkViola I am working in an edit from earlier since I saw my error.
– Rebellos
Dec 18 '18 at 18:14
@MarkViola I was editing before I even saw your solution or comment and rushed to assure you that I was already correcting my answer. I was operating via cellphone initially.
– Rebellos
Dec 18 '18 at 18:21
I understand. I've posted on MSE using a "not-so-smart-phone" many a time and it is definitely a challenge.
– Mark Viola
Dec 18 '18 at 18:24
That is not the LS around $z=1$. You have the Taylor series around $z=0$, for $f$ for $|z|<1$. See my posted solution for the LS around $z=1$.
– Mark Viola
Dec 18 '18 at 18:13
That is not the LS around $z=1$. You have the Taylor series around $z=0$, for $f$ for $|z|<1$. See my posted solution for the LS around $z=1$.
– Mark Viola
Dec 18 '18 at 18:13
@MarkViola I am working in an edit from earlier since I saw my error.
– Rebellos
Dec 18 '18 at 18:14
@MarkViola I am working in an edit from earlier since I saw my error.
– Rebellos
Dec 18 '18 at 18:14
@MarkViola I was editing before I even saw your solution or comment and rushed to assure you that I was already correcting my answer. I was operating via cellphone initially.
– Rebellos
Dec 18 '18 at 18:21
@MarkViola I was editing before I even saw your solution or comment and rushed to assure you that I was already correcting my answer. I was operating via cellphone initially.
– Rebellos
Dec 18 '18 at 18:21
I understand. I've posted on MSE using a "not-so-smart-phone" many a time and it is definitely a challenge.
– Mark Viola
Dec 18 '18 at 18:24
I understand. I've posted on MSE using a "not-so-smart-phone" many a time and it is definitely a challenge.
– Mark Viola
Dec 18 '18 at 18:24
add a comment |
In the annulus $1<|z|<infty$, we have
$$begin{align}
frac{1}{z(1-z)}&=frac{1}{z}+frac1{1-z}\\
&=frac{1}{1+(z-1)}+frac1{1-z}\\
&=frac1{z-1}frac{1}{1+frac1{z-1}}-frac1{z-1}\\
&=frac1{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n-frac1{z-1}\\
&=sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}
end{align}$$
answered Dec 18 '18 at 18:10
Mark Viola
130k1274170
add a comment |
In the annulus $1<|z|<infty$, we have
$$begin{align}
frac{1}{z(1-z)}&=frac{1}{z}+frac1{1-z}\\
&=frac{1}{1+(z-1)}+frac1{1-z}\\
&=frac1{z-1}frac{1}{1+frac1{z-1}}-frac1{z-1}\\
&=frac1{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n-frac1{z-1}\\
&=sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}
end{align}$$
answered Dec 18 '18 at 18:10
Mark Viola
130k1274170
add a comment |
In the annulus $1<|z|<infty$, we have
$$begin{align}
frac{1}{z(1-z)}&=frac{1}{z}+frac1{1-z}\\
&=frac{1}{1+(z-1)}+frac1{1-z}\\
&=frac1{z-1}frac{1}{1+frac1{z-1}}-frac1{z-1}\\
&=frac1{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n-frac1{z-1}\\
&=sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}
end{align}$$
answered Dec 18 '18 at 18:10
Mark Viola
130k1274170
In the annulus $1<|z|<infty$, we have
$$begin{align}
frac{1}{z(1-z)}&=frac{1}{z}+frac1{1-z}\\
&=frac{1}{1+(z-1)}+frac1{1-z}\\
&=frac1{z-1}frac{1}{1+frac1{z-1}}-frac1{z-1}\\
&=frac1{z-1}sum_{n=0}^infty (-1)^n left(frac{1}{z-1}right)^n-frac1{z-1}\\
&=sum_{n=1}^infty (-1)^{n-1}left(frac1{z-1}right)^{n+1}
end{align}$$
answered Dec 18 '18 at 18:10
Mark Viola
130k1274170
answered Dec 18 '18 at 18:10
Mark Viola
130k1274170
answered Dec 18 '18 at 18:10
Mark Viola
130k1274170
answered Dec 18 '18 at 18:10
Mark Viola
130k1274170
130k1274170
add a comment |
add a comment |
We have
$$
eqalign{
& {1 over {zleft( {1 - z} right)}} = cr
& = left{ matrix{
- left( {{1 over z} + {1 over {left( {1 - z} right)}}} right)quad
Rightarrow quad - {1 over z} - sumlimits_{0, le ,n} {z^{,n} } quad left| {,z to 0} right. hfill cr
{1 over {left( {z - 1} right)}} - {1 over {left( {1 + left( {z - 1} right)} right)}}quad Rightarrow quad {1 over {left( {z - 1} right)}}
- sumlimits_{0, le ,n} {left( { - 1} right)^{,n} left( {z - 1} right)^{,n} } quad left| {,z to 1} right. hfill cr
- left( {{1 over z}} right)left( {1 - {1 over {left( {1 - {1 over z}} right)}}} right)quad
Rightarrow quad sumlimits_{0, le ,n} {left( {{1 over z}} right)^{,n + 2} } quad left| {,z to infty } right. hfill cr} right. cr}
$$
answered Dec 18 '18 at 18:28
G Cab
17.9k31237
add a comment |
We have
$$
eqalign{
& {1 over {zleft( {1 - z} right)}} = cr
& = left{ matrix{
- left( {{1 over z} + {1 over {left( {1 - z} right)}}} right)quad
Rightarrow quad - {1 over z} - sumlimits_{0, le ,n} {z^{,n} } quad left| {,z to 0} right. hfill cr
{1 over {left( {z - 1} right)}} - {1 over {left( {1 + left( {z - 1} right)} right)}}quad Rightarrow quad {1 over {left( {z - 1} right)}}
- sumlimits_{0, le ,n} {left( { - 1} right)^{,n} left( {z - 1} right)^{,n} } quad left| {,z to 1} right. hfill cr
- left( {{1 over z}} right)left( {1 - {1 over {left( {1 - {1 over z}} right)}}} right)quad
Rightarrow quad sumlimits_{0, le ,n} {left( {{1 over z}} right)^{,n + 2} } quad left| {,z to infty } right. hfill cr} right. cr}
$$
answered Dec 18 '18 at 18:28
G Cab
17.9k31237
add a comment |
We have
$$
eqalign{
& {1 over {zleft( {1 - z} right)}} = cr
& = left{ matrix{
- left( {{1 over z} + {1 over {left( {1 - z} right)}}} right)quad
Rightarrow quad - {1 over z} - sumlimits_{0, le ,n} {z^{,n} } quad left| {,z to 0} right. hfill cr
{1 over {left( {z - 1} right)}} - {1 over {left( {1 + left( {z - 1} right)} right)}}quad Rightarrow quad {1 over {left( {z - 1} right)}}
- sumlimits_{0, le ,n} {left( { - 1} right)^{,n} left( {z - 1} right)^{,n} } quad left| {,z to 1} right. hfill cr
- left( {{1 over z}} right)left( {1 - {1 over {left( {1 - {1 over z}} right)}}} right)quad
Rightarrow quad sumlimits_{0, le ,n} {left( {{1 over z}} right)^{,n + 2} } quad left| {,z to infty } right. hfill cr} right. cr}
$$
answered Dec 18 '18 at 18:28
G Cab
17.9k31237
We have
$$
eqalign{
& {1 over {zleft( {1 - z} right)}} = cr
& = left{ matrix{
- left( {{1 over z} + {1 over {left( {1 - z} right)}}} right)quad
Rightarrow quad - {1 over z} - sumlimits_{0, le ,n} {z^{,n} } quad left| {,z to 0} right. hfill cr
{1 over {left( {z - 1} right)}} - {1 over {left( {1 + left( {z - 1} right)} right)}}quad Rightarrow quad {1 over {left( {z - 1} right)}}
- sumlimits_{0, le ,n} {left( { - 1} right)^{,n} left( {z - 1} right)^{,n} } quad left| {,z to 1} right. hfill cr
- left( {{1 over z}} right)left( {1 - {1 over {left( {1 - {1 over z}} right)}}} right)quad
Rightarrow quad sumlimits_{0, le ,n} {left( {{1 over z}} right)^{,n + 2} } quad left| {,z to infty } right. hfill cr} right. cr}
$$
answered Dec 18 '18 at 18:28
G Cab
17.9k31237
answered Dec 18 '18 at 18:28
G Cab
17.9k31237
answered Dec 18 '18 at 18:28
G Cab
17.9k31237
answered Dec 18 '18 at 18:28
G Cab
17.9k31237
17.9k31237
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