Double checking a coefficient for a Laurent series, the series for $f(z) = frac{z^2 + 1}{(z - i)^2}dz$
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So we seek the coefficient $a_{-1}$ of the Laurent series for $f(z) = frac{z^2 + 1}{(z - i)^2}dz$, where Laurent series is denoted by
$$f(z) = sum_{n = -infty}^infty a_n(z - z_0)^n$$
when the series is given about the point $z_0 = i$.
We know the coefficients of the Laurent series can be given by the integral
$$a_n = frac{1}{2pi i} int_C frac{f(z)}{(z - z_0)^{n+1}}dz =frac{1}{2pi i} int_C frac{z^2 + 1}{(z-i)^{n+3}}dz$$
where $C$ is a counterclockwise closed curve about $z_0 = i$. Here, the integrand doesn't have any other poles/singularities, so this curve need no other restrictions other than having nonzero radius.
Since we seek $a_{-1}$, we plug in $n = -1$:
$$a_{-1} = frac{1}{2pi i} int_C frac{z^2 + 1}{(z-i)^{2}}dz$$
Given the form of the integral matches the below,
$$frac{1}{2pi i} int_C frac{g(z)}{(z - z_0)^{m+1}}dz=frac{1}{m!}g^{(m)}(z_0)$$
when $g(z)=z^2+1$, $z_0 = i$, and $m=1>0$, we can use this to calculate our integral, yielding
$$a_{-1} = frac{1}{2pi i} int_C frac{z^2 + 1}{(z-i)^{2}}dz= frac{1}{1!}g'(i)=g'(i)$$
Since $g'(z)=2z$, then,
$$a_{-1} = 2i$$
I mostly just wanted to double-check this solution since I feel pretty weak when it comes to Laurent series.
complex-analysis proof-verification laurent-series
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up vote
1
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So we seek the coefficient $a_{-1}$ of the Laurent series for $f(z) = frac{z^2 + 1}{(z - i)^2}dz$, where Laurent series is denoted by
$$f(z) = sum_{n = -infty}^infty a_n(z - z_0)^n$$
when the series is given about the point $z_0 = i$.
We know the coefficients of the Laurent series can be given by the integral
$$a_n = frac{1}{2pi i} int_C frac{f(z)}{(z - z_0)^{n+1}}dz =frac{1}{2pi i} int_C frac{z^2 + 1}{(z-i)^{n+3}}dz$$
where $C$ is a counterclockwise closed curve about $z_0 = i$. Here, the integrand doesn't have any other poles/singularities, so this curve need no other restrictions other than having nonzero radius.
Since we seek $a_{-1}$, we plug in $n = -1$:
$$a_{-1} = frac{1}{2pi i} int_C frac{z^2 + 1}{(z-i)^{2}}dz$$
Given the form of the integral matches the below,
$$frac{1}{2pi i} int_C frac{g(z)}{(z - z_0)^{m+1}}dz=frac{1}{m!}g^{(m)}(z_0)$$
when $g(z)=z^2+1$, $z_0 = i$, and $m=1>0$, we can use this to calculate our integral, yielding
$$a_{-1} = frac{1}{2pi i} int_C frac{z^2 + 1}{(z-i)^{2}}dz= frac{1}{1!}g'(i)=g'(i)$$
Since $g'(z)=2z$, then,
$$a_{-1} = 2i$$
I mostly just wanted to double-check this solution since I feel pretty weak when it comes to Laurent series.
complex-analysis proof-verification laurent-series
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
So we seek the coefficient $a_{-1}$ of the Laurent series for $f(z) = frac{z^2 + 1}{(z - i)^2}dz$, where Laurent series is denoted by
$$f(z) = sum_{n = -infty}^infty a_n(z - z_0)^n$$
when the series is given about the point $z_0 = i$.
We know the coefficients of the Laurent series can be given by the integral
$$a_n = frac{1}{2pi i} int_C frac{f(z)}{(z - z_0)^{n+1}}dz =frac{1}{2pi i} int_C frac{z^2 + 1}{(z-i)^{n+3}}dz$$
where $C$ is a counterclockwise closed curve about $z_0 = i$. Here, the integrand doesn't have any other poles/singularities, so this curve need no other restrictions other than having nonzero radius.
Since we seek $a_{-1}$, we plug in $n = -1$:
$$a_{-1} = frac{1}{2pi i} int_C frac{z^2 + 1}{(z-i)^{2}}dz$$
Given the form of the integral matches the below,
$$frac{1}{2pi i} int_C frac{g(z)}{(z - z_0)^{m+1}}dz=frac{1}{m!}g^{(m)}(z_0)$$
when $g(z)=z^2+1$, $z_0 = i$, and $m=1>0$, we can use this to calculate our integral, yielding
$$a_{-1} = frac{1}{2pi i} int_C frac{z^2 + 1}{(z-i)^{2}}dz= frac{1}{1!}g'(i)=g'(i)$$
Since $g'(z)=2z$, then,
$$a_{-1} = 2i$$
I mostly just wanted to double-check this solution since I feel pretty weak when it comes to Laurent series.
complex-analysis proof-verification laurent-series
So we seek the coefficient $a_{-1}$ of the Laurent series for $f(z) = frac{z^2 + 1}{(z - i)^2}dz$, where Laurent series is denoted by
$$f(z) = sum_{n = -infty}^infty a_n(z - z_0)^n$$
when the series is given about the point $z_0 = i$.
We know the coefficients of the Laurent series can be given by the integral
$$a_n = frac{1}{2pi i} int_C frac{f(z)}{(z - z_0)^{n+1}}dz =frac{1}{2pi i} int_C frac{z^2 + 1}{(z-i)^{n+3}}dz$$
where $C$ is a counterclockwise closed curve about $z_0 = i$. Here, the integrand doesn't have any other poles/singularities, so this curve need no other restrictions other than having nonzero radius.
Since we seek $a_{-1}$, we plug in $n = -1$:
$$a_{-1} = frac{1}{2pi i} int_C frac{z^2 + 1}{(z-i)^{2}}dz$$
Given the form of the integral matches the below,
$$frac{1}{2pi i} int_C frac{g(z)}{(z - z_0)^{m+1}}dz=frac{1}{m!}g^{(m)}(z_0)$$
when $g(z)=z^2+1$, $z_0 = i$, and $m=1>0$, we can use this to calculate our integral, yielding
$$a_{-1} = frac{1}{2pi i} int_C frac{z^2 + 1}{(z-i)^{2}}dz= frac{1}{1!}g'(i)=g'(i)$$
Since $g'(z)=2z$, then,
$$a_{-1} = 2i$$
I mostly just wanted to double-check this solution since I feel pretty weak when it comes to Laurent series.
complex-analysis proof-verification laurent-series
complex-analysis proof-verification laurent-series
asked Nov 14 at 4:37
Eevee Trainer
7938
7938
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2 Answers
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Observe: $f(z) = frac{z^2+1}{(z-i)^2} = frac{(z+i)(z-i)}{(z-i)^2} = frac{z+i}{z-i} = frac{(z-i) +i +i}{z-i} = frac{2i}{z-i} + 1.$ What is the coefficient for $a_{-1}$ given this observation?
I believe that, following from the definition of Laurent series, we would have $2i$ as well per this derivation, since $$frac{2i}{z-i}=2i(z-i)^{-1}$$ with the power of $(z-i)$ hinting that this refers to the $a_{-1}$ term. And thus, similarly, if I wanted to find $a_0$, $a_0 = 1$. In which case I'd be correct, right? Just want to be sure.
– Eevee Trainer
Nov 14 at 5:05
1
@EeveeTrainer indeed!
– user328442
Nov 14 at 5:06
add a comment |
up vote
1
down vote
Just wanted to give some context to the already accepted answer:
In practice, when trying to compute the Laurent series for a complex function, we (almost) never use the definitions provided by the Laurent theorem to calculate the coefficients. Rather, we fool around with the Taylor series of known portions of the function until we get one sum with possibly negative powers of $z$, as in the accepted answer. This typically involves tricks like multiplying by 1, adding 0, and things like that. This in turn gives information about the integrals, according to the definitions
$$
a_n = frac{1}{2pi i}oint_{+gamma}frac{f(w)}{(w-z_0)^{n+1}}dw,
b_n = frac{1}{2pi i}oint_{+gamma}f(w)(w-z_0)^{n-1}dw
$$
where $gamma$ is any curve defined in the annulus on which you are computing the Laurent series. This is well defined by the homotopy principle.
So in this case, we didn't need any Taylor series expansions, the expression happened to be nice enough to simplify to a Laurent series already.
New contributor
Yeah I get that, it's still a mode of thought I'm trying to get into myself though since my professor waffles on the idea. Sometimes he seems to want the formal calculations of the coefficients, other times he wants us to do it the easy way, etc. I figured he specifically wanted the formal calculation in this case given a comment earlier in the evening (mostly just trying to contextualize this question/comment), though the alternate method was also helpful and illustrative as well.
– Eevee Trainer
Nov 14 at 5:33
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Observe: $f(z) = frac{z^2+1}{(z-i)^2} = frac{(z+i)(z-i)}{(z-i)^2} = frac{z+i}{z-i} = frac{(z-i) +i +i}{z-i} = frac{2i}{z-i} + 1.$ What is the coefficient for $a_{-1}$ given this observation?
I believe that, following from the definition of Laurent series, we would have $2i$ as well per this derivation, since $$frac{2i}{z-i}=2i(z-i)^{-1}$$ with the power of $(z-i)$ hinting that this refers to the $a_{-1}$ term. And thus, similarly, if I wanted to find $a_0$, $a_0 = 1$. In which case I'd be correct, right? Just want to be sure.
– Eevee Trainer
Nov 14 at 5:05
1
@EeveeTrainer indeed!
– user328442
Nov 14 at 5:06
add a comment |
up vote
1
down vote
accepted
Observe: $f(z) = frac{z^2+1}{(z-i)^2} = frac{(z+i)(z-i)}{(z-i)^2} = frac{z+i}{z-i} = frac{(z-i) +i +i}{z-i} = frac{2i}{z-i} + 1.$ What is the coefficient for $a_{-1}$ given this observation?
I believe that, following from the definition of Laurent series, we would have $2i$ as well per this derivation, since $$frac{2i}{z-i}=2i(z-i)^{-1}$$ with the power of $(z-i)$ hinting that this refers to the $a_{-1}$ term. And thus, similarly, if I wanted to find $a_0$, $a_0 = 1$. In which case I'd be correct, right? Just want to be sure.
– Eevee Trainer
Nov 14 at 5:05
1
@EeveeTrainer indeed!
– user328442
Nov 14 at 5:06
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Observe: $f(z) = frac{z^2+1}{(z-i)^2} = frac{(z+i)(z-i)}{(z-i)^2} = frac{z+i}{z-i} = frac{(z-i) +i +i}{z-i} = frac{2i}{z-i} + 1.$ What is the coefficient for $a_{-1}$ given this observation?
Observe: $f(z) = frac{z^2+1}{(z-i)^2} = frac{(z+i)(z-i)}{(z-i)^2} = frac{z+i}{z-i} = frac{(z-i) +i +i}{z-i} = frac{2i}{z-i} + 1.$ What is the coefficient for $a_{-1}$ given this observation?
answered Nov 14 at 5:01
user328442
1,7881516
1,7881516
I believe that, following from the definition of Laurent series, we would have $2i$ as well per this derivation, since $$frac{2i}{z-i}=2i(z-i)^{-1}$$ with the power of $(z-i)$ hinting that this refers to the $a_{-1}$ term. And thus, similarly, if I wanted to find $a_0$, $a_0 = 1$. In which case I'd be correct, right? Just want to be sure.
– Eevee Trainer
Nov 14 at 5:05
1
@EeveeTrainer indeed!
– user328442
Nov 14 at 5:06
add a comment |
I believe that, following from the definition of Laurent series, we would have $2i$ as well per this derivation, since $$frac{2i}{z-i}=2i(z-i)^{-1}$$ with the power of $(z-i)$ hinting that this refers to the $a_{-1}$ term. And thus, similarly, if I wanted to find $a_0$, $a_0 = 1$. In which case I'd be correct, right? Just want to be sure.
– Eevee Trainer
Nov 14 at 5:05
1
@EeveeTrainer indeed!
– user328442
Nov 14 at 5:06
I believe that, following from the definition of Laurent series, we would have $2i$ as well per this derivation, since $$frac{2i}{z-i}=2i(z-i)^{-1}$$ with the power of $(z-i)$ hinting that this refers to the $a_{-1}$ term. And thus, similarly, if I wanted to find $a_0$, $a_0 = 1$. In which case I'd be correct, right? Just want to be sure.
– Eevee Trainer
Nov 14 at 5:05
I believe that, following from the definition of Laurent series, we would have $2i$ as well per this derivation, since $$frac{2i}{z-i}=2i(z-i)^{-1}$$ with the power of $(z-i)$ hinting that this refers to the $a_{-1}$ term. And thus, similarly, if I wanted to find $a_0$, $a_0 = 1$. In which case I'd be correct, right? Just want to be sure.
– Eevee Trainer
Nov 14 at 5:05
1
1
@EeveeTrainer indeed!
– user328442
Nov 14 at 5:06
@EeveeTrainer indeed!
– user328442
Nov 14 at 5:06
add a comment |
up vote
1
down vote
Just wanted to give some context to the already accepted answer:
In practice, when trying to compute the Laurent series for a complex function, we (almost) never use the definitions provided by the Laurent theorem to calculate the coefficients. Rather, we fool around with the Taylor series of known portions of the function until we get one sum with possibly negative powers of $z$, as in the accepted answer. This typically involves tricks like multiplying by 1, adding 0, and things like that. This in turn gives information about the integrals, according to the definitions
$$
a_n = frac{1}{2pi i}oint_{+gamma}frac{f(w)}{(w-z_0)^{n+1}}dw,
b_n = frac{1}{2pi i}oint_{+gamma}f(w)(w-z_0)^{n-1}dw
$$
where $gamma$ is any curve defined in the annulus on which you are computing the Laurent series. This is well defined by the homotopy principle.
So in this case, we didn't need any Taylor series expansions, the expression happened to be nice enough to simplify to a Laurent series already.
New contributor
Yeah I get that, it's still a mode of thought I'm trying to get into myself though since my professor waffles on the idea. Sometimes he seems to want the formal calculations of the coefficients, other times he wants us to do it the easy way, etc. I figured he specifically wanted the formal calculation in this case given a comment earlier in the evening (mostly just trying to contextualize this question/comment), though the alternate method was also helpful and illustrative as well.
– Eevee Trainer
Nov 14 at 5:33
add a comment |
up vote
1
down vote
Just wanted to give some context to the already accepted answer:
In practice, when trying to compute the Laurent series for a complex function, we (almost) never use the definitions provided by the Laurent theorem to calculate the coefficients. Rather, we fool around with the Taylor series of known portions of the function until we get one sum with possibly negative powers of $z$, as in the accepted answer. This typically involves tricks like multiplying by 1, adding 0, and things like that. This in turn gives information about the integrals, according to the definitions
$$
a_n = frac{1}{2pi i}oint_{+gamma}frac{f(w)}{(w-z_0)^{n+1}}dw,
b_n = frac{1}{2pi i}oint_{+gamma}f(w)(w-z_0)^{n-1}dw
$$
where $gamma$ is any curve defined in the annulus on which you are computing the Laurent series. This is well defined by the homotopy principle.
So in this case, we didn't need any Taylor series expansions, the expression happened to be nice enough to simplify to a Laurent series already.
New contributor
Yeah I get that, it's still a mode of thought I'm trying to get into myself though since my professor waffles on the idea. Sometimes he seems to want the formal calculations of the coefficients, other times he wants us to do it the easy way, etc. I figured he specifically wanted the formal calculation in this case given a comment earlier in the evening (mostly just trying to contextualize this question/comment), though the alternate method was also helpful and illustrative as well.
– Eevee Trainer
Nov 14 at 5:33
add a comment |
up vote
1
down vote
up vote
1
down vote
Just wanted to give some context to the already accepted answer:
In practice, when trying to compute the Laurent series for a complex function, we (almost) never use the definitions provided by the Laurent theorem to calculate the coefficients. Rather, we fool around with the Taylor series of known portions of the function until we get one sum with possibly negative powers of $z$, as in the accepted answer. This typically involves tricks like multiplying by 1, adding 0, and things like that. This in turn gives information about the integrals, according to the definitions
$$
a_n = frac{1}{2pi i}oint_{+gamma}frac{f(w)}{(w-z_0)^{n+1}}dw,
b_n = frac{1}{2pi i}oint_{+gamma}f(w)(w-z_0)^{n-1}dw
$$
where $gamma$ is any curve defined in the annulus on which you are computing the Laurent series. This is well defined by the homotopy principle.
So in this case, we didn't need any Taylor series expansions, the expression happened to be nice enough to simplify to a Laurent series already.
New contributor
Just wanted to give some context to the already accepted answer:
In practice, when trying to compute the Laurent series for a complex function, we (almost) never use the definitions provided by the Laurent theorem to calculate the coefficients. Rather, we fool around with the Taylor series of known portions of the function until we get one sum with possibly negative powers of $z$, as in the accepted answer. This typically involves tricks like multiplying by 1, adding 0, and things like that. This in turn gives information about the integrals, according to the definitions
$$
a_n = frac{1}{2pi i}oint_{+gamma}frac{f(w)}{(w-z_0)^{n+1}}dw,
b_n = frac{1}{2pi i}oint_{+gamma}f(w)(w-z_0)^{n-1}dw
$$
where $gamma$ is any curve defined in the annulus on which you are computing the Laurent series. This is well defined by the homotopy principle.
So in this case, we didn't need any Taylor series expansions, the expression happened to be nice enough to simplify to a Laurent series already.
New contributor
New contributor
answered Nov 14 at 5:28
staedtlerr
413
413
New contributor
New contributor
Yeah I get that, it's still a mode of thought I'm trying to get into myself though since my professor waffles on the idea. Sometimes he seems to want the formal calculations of the coefficients, other times he wants us to do it the easy way, etc. I figured he specifically wanted the formal calculation in this case given a comment earlier in the evening (mostly just trying to contextualize this question/comment), though the alternate method was also helpful and illustrative as well.
– Eevee Trainer
Nov 14 at 5:33
add a comment |
Yeah I get that, it's still a mode of thought I'm trying to get into myself though since my professor waffles on the idea. Sometimes he seems to want the formal calculations of the coefficients, other times he wants us to do it the easy way, etc. I figured he specifically wanted the formal calculation in this case given a comment earlier in the evening (mostly just trying to contextualize this question/comment), though the alternate method was also helpful and illustrative as well.
– Eevee Trainer
Nov 14 at 5:33
Yeah I get that, it's still a mode of thought I'm trying to get into myself though since my professor waffles on the idea. Sometimes he seems to want the formal calculations of the coefficients, other times he wants us to do it the easy way, etc. I figured he specifically wanted the formal calculation in this case given a comment earlier in the evening (mostly just trying to contextualize this question/comment), though the alternate method was also helpful and illustrative as well.
– Eevee Trainer
Nov 14 at 5:33
Yeah I get that, it's still a mode of thought I'm trying to get into myself though since my professor waffles on the idea. Sometimes he seems to want the formal calculations of the coefficients, other times he wants us to do it the easy way, etc. I figured he specifically wanted the formal calculation in this case given a comment earlier in the evening (mostly just trying to contextualize this question/comment), though the alternate method was also helpful and illustrative as well.
– Eevee Trainer
Nov 14 at 5:33
add a comment |
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