Find residues of $f(z)=frac{1}{(e^{z}-1)^{2}}$
up vote
3
down vote
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How to find the residues of $f(z)=dfrac{1}{(e^{z}-1)^{2}}$
I have found that the poles $z=2pi i n$.
But when I apply the formula $dfrac{1}{(m-1)!}limlimits_{zrightarrow z_{0}}dfrac{d^{m-1}}{dz^{m-1}}left((z-z_{0})^{m}f(z)right)$,
I got $limlimits_{zrightarrow z_{0}}dfrac{d}{dz}(z-z_{0})^{2}dfrac{1}{(e^{z}-1)^{2}}$,
which is so complicated that I cannot get the limit value.
complex-analysis residue-calculus
edited Nov 23 at 17:39
Martin Sleziak
44.5k7115268
asked Nov 23 at 12:50
Ganlin Zh
184
add a comment |
up vote
3
down vote
favorite
How to find the residues of $f(z)=dfrac{1}{(e^{z}-1)^{2}}$
I have found that the poles $z=2pi i n$.
But when I apply the formula $dfrac{1}{(m-1)!}limlimits_{zrightarrow z_{0}}dfrac{d^{m-1}}{dz^{m-1}}left((z-z_{0})^{m}f(z)right)$,
I got $limlimits_{zrightarrow z_{0}}dfrac{d}{dz}(z-z_{0})^{2}dfrac{1}{(e^{z}-1)^{2}}$,
which is so complicated that I cannot get the limit value.
complex-analysis residue-calculus
edited Nov 23 at 17:39
Martin Sleziak
44.5k7115268
asked Nov 23 at 12:50
Ganlin Zh
184
1
Apply L'Hospital to your limit...It really isn't that complicated
– DonAntonio
Nov 23 at 13:26
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
How to find the residues of $f(z)=dfrac{1}{(e^{z}-1)^{2}}$
I have found that the poles $z=2pi i n$.
But when I apply the formula $dfrac{1}{(m-1)!}limlimits_{zrightarrow z_{0}}dfrac{d^{m-1}}{dz^{m-1}}left((z-z_{0})^{m}f(z)right)$,
I got $limlimits_{zrightarrow z_{0}}dfrac{d}{dz}(z-z_{0})^{2}dfrac{1}{(e^{z}-1)^{2}}$,
which is so complicated that I cannot get the limit value.
complex-analysis residue-calculus
edited Nov 23 at 17:39
Martin Sleziak
44.5k7115268
asked Nov 23 at 12:50
Ganlin Zh
184
How to find the residues of $f(z)=dfrac{1}{(e^{z}-1)^{2}}$
I have found that the poles $z=2pi i n$.
But when I apply the formula $dfrac{1}{(m-1)!}limlimits_{zrightarrow z_{0}}dfrac{d^{m-1}}{dz^{m-1}}left((z-z_{0})^{m}f(z)right)$,
I got $limlimits_{zrightarrow z_{0}}dfrac{d}{dz}(z-z_{0})^{2}dfrac{1}{(e^{z}-1)^{2}}$,
which is so complicated that I cannot get the limit value.
complex-analysis residue-calculus
complex-analysis residue-calculus
edited Nov 23 at 17:39
Martin Sleziak
44.5k7115268
asked Nov 23 at 12:50
Ganlin Zh
184
edited Nov 23 at 17:39
Martin Sleziak
44.5k7115268
asked Nov 23 at 12:50
Ganlin Zh
184
edited Nov 23 at 17:39
Martin Sleziak
44.5k7115268
edited Nov 23 at 17:39
Martin Sleziak
44.5k7115268
edited Nov 23 at 17:39
Martin Sleziak
44.5k7115268
44.5k7115268
asked Nov 23 at 12:50
Ganlin Zh
184
asked Nov 23 at 12:50
Ganlin Zh
184
asked Nov 23 at 12:50
Ganlin Zh
184
184
1
Apply L'Hospital to your limit...It really isn't that complicated
– DonAntonio
Nov 23 at 13:26
add a comment |
1
Apply L'Hospital to your limit...It really isn't that complicated
– DonAntonio
Nov 23 at 13:26
1
1
Apply L'Hospital to your limit...It really isn't that complicated
– DonAntonio
Nov 23 at 13:26
Apply L'Hospital to your limit...It really isn't that complicated
– DonAntonio
Nov 23 at 13:26
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Hint:
This function is periodic with period $2pi i$, therefore all residues on $z=2pi i n$ are equal. Then it's sufficient to calculate residues at $z=0$:
Edit:
$$lim_{zto0}dfrac{d}{dz}z^2dfrac{1}{(e^{z}-1)^2}=lim_{zto0}frac{2zleft(e^z-1-ze^zright)}{(e^z-1)^3}=color{blue}{-1}$$
answered Nov 23 at 12:58
Nosrati
26.2k62353
thanks a lot, solved
– Ganlin Zh
Nov 23 at 13:02
OK. You are welcome. $:)$
– Nosrati
Nov 23 at 13:03
add a comment |
up vote
2
down vote
Many time it is difficult to apply usual derivative as we get repetative terms .So in such case if we know series of that function it will be easy.
$frac{1}{(1+z+z^2/2+o(z^3)-1)^2}$=$frac{1}{z^2(1+z/2)^2}$=$frac{1-z+o(z^2)}{z^2}$
By defination residue is coefficent of 1/z implies -1.
answered Nov 24 at 3:34
Shubham
1,5821519
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint:
This function is periodic with period $2pi i$, therefore all residues on $z=2pi i n$ are equal. Then it's sufficient to calculate residues at $z=0$:
Edit:
$$lim_{zto0}dfrac{d}{dz}z^2dfrac{1}{(e^{z}-1)^2}=lim_{zto0}frac{2zleft(e^z-1-ze^zright)}{(e^z-1)^3}=color{blue}{-1}$$
answered Nov 23 at 12:58
Nosrati
26.2k62353
thanks a lot, solved
– Ganlin Zh
Nov 23 at 13:02
OK. You are welcome. $:)$
– Nosrati
Nov 23 at 13:03
add a comment |
up vote
2
down vote
accepted
Hint:
This function is periodic with period $2pi i$, therefore all residues on $z=2pi i n$ are equal. Then it's sufficient to calculate residues at $z=0$:
Edit:
$$lim_{zto0}dfrac{d}{dz}z^2dfrac{1}{(e^{z}-1)^2}=lim_{zto0}frac{2zleft(e^z-1-ze^zright)}{(e^z-1)^3}=color{blue}{-1}$$
answered Nov 23 at 12:58
Nosrati
26.2k62353
thanks a lot, solved
– Ganlin Zh
Nov 23 at 13:02
OK. You are welcome. $:)$
– Nosrati
Nov 23 at 13:03
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint:
This function is periodic with period $2pi i$, therefore all residues on $z=2pi i n$ are equal. Then it's sufficient to calculate residues at $z=0$:
Edit:
$$lim_{zto0}dfrac{d}{dz}z^2dfrac{1}{(e^{z}-1)^2}=lim_{zto0}frac{2zleft(e^z-1-ze^zright)}{(e^z-1)^3}=color{blue}{-1}$$
answered Nov 23 at 12:58
Nosrati
26.2k62353
Hint:
This function is periodic with period $2pi i$, therefore all residues on $z=2pi i n$ are equal. Then it's sufficient to calculate residues at $z=0$:
Edit:
$$lim_{zto0}dfrac{d}{dz}z^2dfrac{1}{(e^{z}-1)^2}=lim_{zto0}frac{2zleft(e^z-1-ze^zright)}{(e^z-1)^3}=color{blue}{-1}$$
answered Nov 23 at 12:58
Nosrati
26.2k62353
edited Nov 23 at 13:54
answered Nov 23 at 12:58
Nosrati
26.2k62353
answered Nov 23 at 12:58
Nosrati
26.2k62353
answered Nov 23 at 12:58
Nosrati
26.2k62353
26.2k62353
thanks a lot, solved
– Ganlin Zh
Nov 23 at 13:02
OK. You are welcome. $:)$
– Nosrati
Nov 23 at 13:03
add a comment |
thanks a lot, solved
– Ganlin Zh
Nov 23 at 13:02
OK. You are welcome. $:)$
– Nosrati
Nov 23 at 13:03
thanks a lot, solved
– Ganlin Zh
Nov 23 at 13:02
thanks a lot, solved
– Ganlin Zh
Nov 23 at 13:02
OK. You are welcome. $:)$
– Nosrati
Nov 23 at 13:03
OK. You are welcome. $:)$
– Nosrati
Nov 23 at 13:03
add a comment |
up vote
2
down vote
Many time it is difficult to apply usual derivative as we get repetative terms .So in such case if we know series of that function it will be easy.
$frac{1}{(1+z+z^2/2+o(z^3)-1)^2}$=$frac{1}{z^2(1+z/2)^2}$=$frac{1-z+o(z^2)}{z^2}$
By defination residue is coefficent of 1/z implies -1.
answered Nov 24 at 3:34
Shubham
1,5821519
add a comment |
up vote
2
down vote
Many time it is difficult to apply usual derivative as we get repetative terms .So in such case if we know series of that function it will be easy.
$frac{1}{(1+z+z^2/2+o(z^3)-1)^2}$=$frac{1}{z^2(1+z/2)^2}$=$frac{1-z+o(z^2)}{z^2}$
By defination residue is coefficent of 1/z implies -1.
answered Nov 24 at 3:34
Shubham
1,5821519
add a comment |
up vote
2
down vote
up vote
2
down vote
Many time it is difficult to apply usual derivative as we get repetative terms .So in such case if we know series of that function it will be easy.
$frac{1}{(1+z+z^2/2+o(z^3)-1)^2}$=$frac{1}{z^2(1+z/2)^2}$=$frac{1-z+o(z^2)}{z^2}$
By defination residue is coefficent of 1/z implies -1.
answered Nov 24 at 3:34
Shubham
1,5821519
Many time it is difficult to apply usual derivative as we get repetative terms .So in such case if we know series of that function it will be easy.
$frac{1}{(1+z+z^2/2+o(z^3)-1)^2}$=$frac{1}{z^2(1+z/2)^2}$=$frac{1-z+o(z^2)}{z^2}$
By defination residue is coefficent of 1/z implies -1.
answered Nov 24 at 3:34
Shubham
1,5821519
answered Nov 24 at 3:34
Shubham
1,5821519
answered Nov 24 at 3:34
Shubham
1,5821519
answered Nov 24 at 3:34
Shubham
1,5821519
1,5821519
add a comment |
add a comment |
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Apply L'Hospital to your limit...It really isn't that complicated
– DonAntonio
Nov 23 at 13:26