Complex functions and residues
I know that, since the complex function
$$f(z)=frac{1}{z(e^z-1)}$$
have a pole of 2nd order in z=0, I should be able to represent it as:
$$f(z)=frac{g(z)}{(z-0)^2}$$
Being g(z) a function that is both analytic and different from zero when z=0.
But the only expression for g(z) that I was able to get to was
$$g(z)=frac{z}{e^z-1}$$
wich doesn't meets the criteria explained before.
Can anyone say what am I getting wrong?
Does f(z) really have a pole of 2nd order in zero?
complex-analysis
asked Dec 10 '18 at 2:14
Rodrigo Castañon
81
|
show 10 more comments
I know that, since the complex function
$$f(z)=frac{1}{z(e^z-1)}$$
have a pole of 2nd order in z=0, I should be able to represent it as:
$$f(z)=frac{g(z)}{(z-0)^2}$$
Being g(z) a function that is both analytic and different from zero when z=0.
But the only expression for g(z) that I was able to get to was
$$g(z)=frac{z}{e^z-1}$$
wich doesn't meets the criteria explained before.
Can anyone say what am I getting wrong?
Does f(z) really have a pole of 2nd order in zero?
complex-analysis
asked Dec 10 '18 at 2:14
Rodrigo Castañon
81
"wich doesn't meets the criteria explained before" Hmmm... why do you think it does not?
– Did
Dec 10 '18 at 2:50
@Did well, because g(z) has a singularity when z=0, doensn´t it? That would make it a non-analytic function in this point from what I know, but I could be wrong of course
– Rodrigo Castañon
Dec 10 '18 at 10:37
$g(z)$ has a removable singularity at $z=0$, since $lim_{zto 0} g(z) = 1$. You can make it analytic by defining $$ g(z) = begin{cases} dfrac{z}{e^z-1}, & z ne 0 \ 1, & z = 0 end{cases} $$
– Dylan
Dec 10 '18 at 14:42
How is the post below that you chose to accept, answering the question?
– Did
Dec 10 '18 at 15:16
@Dylan yes, I realized that after a while, thank you!
– Rodrigo Castañon
Dec 10 '18 at 17:51
|
show 10 more comments
I know that, since the complex function
$$f(z)=frac{1}{z(e^z-1)}$$
have a pole of 2nd order in z=0, I should be able to represent it as:
$$f(z)=frac{g(z)}{(z-0)^2}$$
Being g(z) a function that is both analytic and different from zero when z=0.
But the only expression for g(z) that I was able to get to was
$$g(z)=frac{z}{e^z-1}$$
wich doesn't meets the criteria explained before.
Can anyone say what am I getting wrong?
Does f(z) really have a pole of 2nd order in zero?
complex-analysis
asked Dec 10 '18 at 2:14
Rodrigo Castañon
81
I know that, since the complex function
$$f(z)=frac{1}{z(e^z-1)}$$
have a pole of 2nd order in z=0, I should be able to represent it as:
$$f(z)=frac{g(z)}{(z-0)^2}$$
Being g(z) a function that is both analytic and different from zero when z=0.
But the only expression for g(z) that I was able to get to was
$$g(z)=frac{z}{e^z-1}$$
wich doesn't meets the criteria explained before.
Can anyone say what am I getting wrong?
Does f(z) really have a pole of 2nd order in zero?
complex-analysis
complex-analysis
asked Dec 10 '18 at 2:14
Rodrigo Castañon
81
asked Dec 10 '18 at 2:14
Rodrigo Castañon
81
asked Dec 10 '18 at 2:14
Rodrigo Castañon
81
asked Dec 10 '18 at 2:14
Rodrigo Castañon
81
asked Dec 10 '18 at 2:14
Rodrigo Castañon
81
81
"wich doesn't meets the criteria explained before" Hmmm... why do you think it does not?
– Did
Dec 10 '18 at 2:50
@Did well, because g(z) has a singularity when z=0, doensn´t it? That would make it a non-analytic function in this point from what I know, but I could be wrong of course
– Rodrigo Castañon
Dec 10 '18 at 10:37
$g(z)$ has a removable singularity at $z=0$, since $lim_{zto 0} g(z) = 1$. You can make it analytic by defining $$ g(z) = begin{cases} dfrac{z}{e^z-1}, & z ne 0 \ 1, & z = 0 end{cases} $$
– Dylan
Dec 10 '18 at 14:42
How is the post below that you chose to accept, answering the question?
– Did
Dec 10 '18 at 15:16
@Dylan yes, I realized that after a while, thank you!
– Rodrigo Castañon
Dec 10 '18 at 17:51
|
show 10 more comments
"wich doesn't meets the criteria explained before" Hmmm... why do you think it does not?
– Did
Dec 10 '18 at 2:50
@Did well, because g(z) has a singularity when z=0, doensn´t it? That would make it a non-analytic function in this point from what I know, but I could be wrong of course
– Rodrigo Castañon
Dec 10 '18 at 10:37
$g(z)$ has a removable singularity at $z=0$, since $lim_{zto 0} g(z) = 1$. You can make it analytic by defining $$ g(z) = begin{cases} dfrac{z}{e^z-1}, & z ne 0 \ 1, & z = 0 end{cases} $$
– Dylan
Dec 10 '18 at 14:42
How is the post below that you chose to accept, answering the question?
– Did
Dec 10 '18 at 15:16
@Dylan yes, I realized that after a while, thank you!
– Rodrigo Castañon
Dec 10 '18 at 17:51
"wich doesn't meets the criteria explained before" Hmmm... why do you think it does not?
– Did
Dec 10 '18 at 2:50
"wich doesn't meets the criteria explained before" Hmmm... why do you think it does not?
– Did
Dec 10 '18 at 2:50
@Did well, because g(z) has a singularity when z=0, doensn´t it? That would make it a non-analytic function in this point from what I know, but I could be wrong of course
– Rodrigo Castañon
Dec 10 '18 at 10:37
@Did well, because g(z) has a singularity when z=0, doensn´t it? That would make it a non-analytic function in this point from what I know, but I could be wrong of course
– Rodrigo Castañon
Dec 10 '18 at 10:37
$g(z)$ has a removable singularity at $z=0$, since $lim_{zto 0} g(z) = 1$. You can make it analytic by defining $$ g(z) = begin{cases} dfrac{z}{e^z-1}, & z ne 0 \ 1, & z = 0 end{cases} $$
– Dylan
Dec 10 '18 at 14:42
$g(z)$ has a removable singularity at $z=0$, since $lim_{zto 0} g(z) = 1$. You can make it analytic by defining $$ g(z) = begin{cases} dfrac{z}{e^z-1}, & z ne 0 \ 1, & z = 0 end{cases} $$
– Dylan
Dec 10 '18 at 14:42
How is the post below that you chose to accept, answering the question?
– Did
Dec 10 '18 at 15:16
How is the post below that you chose to accept, answering the question?
– Did
Dec 10 '18 at 15:16
@Dylan yes, I realized that after a while, thank you!
– Rodrigo Castañon
Dec 10 '18 at 17:51
@Dylan yes, I realized that after a while, thank you!
– Rodrigo Castañon
Dec 10 '18 at 17:51
|
show 10 more comments
1 Answer
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I'm just taking my first complex analysis but from my understanding you are correct in the fact that pole z= 0 is of second order.
Using the formula for higher order poles.
$$Res_{z rightarrow x} = frac{1}{(n-1)!} lim_{zrightarrow x} frac{d^{n-1}}{dz^{n-1}}(z-x)^n f(z)$$
Where x is the pole 0 and n is the order.
Using this formula:
$$Res_{z rightarrow 0} = frac{1}{(2-1)!} lim_{zrightarrow 0} frac{d^{2-1}}{dz^{2-1}}(z-0)^2 f(z)$$
$$Res_{z rightarrow 0} = frac{1}{(1)!} lim_{zrightarrow x} frac{d}{dz}(z-0)^2 frac{1}{z(e^z - 1)}$$
$$ lim_{zrightarrow 0} frac{d}{dz}(z)^2 frac{1}{z(e^z - 1)}$$
$$ lim_{zrightarrow 0} frac{d}{dz}frac{z }{(e^z - 1)}$$
$$ lim_{zrightarrow 0} frac{(e^z - 1) - (e^z)z}{(e^z - 1)^2}$$
Now just do a couple L'Hopitals and you should get:
Res at z=0:
$$= - frac{1}{2}$$
answered Dec 10 '18 at 2:38
Safder
15110
1
Thank you! I've just started to study the subject too and your thoughts on the question were very helpful :)
– Rodrigo Castañon
Dec 10 '18 at 11:06
Glad to be of help! @RodrigoCastañon
– Safder
Dec 10 '18 at 22:32
add a comment |
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I'm just taking my first complex analysis but from my understanding you are correct in the fact that pole z= 0 is of second order.
Using the formula for higher order poles.
$$Res_{z rightarrow x} = frac{1}{(n-1)!} lim_{zrightarrow x} frac{d^{n-1}}{dz^{n-1}}(z-x)^n f(z)$$
Where x is the pole 0 and n is the order.
Using this formula:
$$Res_{z rightarrow 0} = frac{1}{(2-1)!} lim_{zrightarrow 0} frac{d^{2-1}}{dz^{2-1}}(z-0)^2 f(z)$$
$$Res_{z rightarrow 0} = frac{1}{(1)!} lim_{zrightarrow x} frac{d}{dz}(z-0)^2 frac{1}{z(e^z - 1)}$$
$$ lim_{zrightarrow 0} frac{d}{dz}(z)^2 frac{1}{z(e^z - 1)}$$
$$ lim_{zrightarrow 0} frac{d}{dz}frac{z }{(e^z - 1)}$$
$$ lim_{zrightarrow 0} frac{(e^z - 1) - (e^z)z}{(e^z - 1)^2}$$
Now just do a couple L'Hopitals and you should get:
Res at z=0:
$$= - frac{1}{2}$$
answered Dec 10 '18 at 2:38
Safder
15110
1
Thank you! I've just started to study the subject too and your thoughts on the question were very helpful :)
– Rodrigo Castañon
Dec 10 '18 at 11:06
Glad to be of help! @RodrigoCastañon
– Safder
Dec 10 '18 at 22:32
add a comment |
I'm just taking my first complex analysis but from my understanding you are correct in the fact that pole z= 0 is of second order.
Using the formula for higher order poles.
$$Res_{z rightarrow x} = frac{1}{(n-1)!} lim_{zrightarrow x} frac{d^{n-1}}{dz^{n-1}}(z-x)^n f(z)$$
Where x is the pole 0 and n is the order.
Using this formula:
$$Res_{z rightarrow 0} = frac{1}{(2-1)!} lim_{zrightarrow 0} frac{d^{2-1}}{dz^{2-1}}(z-0)^2 f(z)$$
$$Res_{z rightarrow 0} = frac{1}{(1)!} lim_{zrightarrow x} frac{d}{dz}(z-0)^2 frac{1}{z(e^z - 1)}$$
$$ lim_{zrightarrow 0} frac{d}{dz}(z)^2 frac{1}{z(e^z - 1)}$$
$$ lim_{zrightarrow 0} frac{d}{dz}frac{z }{(e^z - 1)}$$
$$ lim_{zrightarrow 0} frac{(e^z - 1) - (e^z)z}{(e^z - 1)^2}$$
Now just do a couple L'Hopitals and you should get:
Res at z=0:
$$= - frac{1}{2}$$
answered Dec 10 '18 at 2:38
Safder
15110
1
Thank you! I've just started to study the subject too and your thoughts on the question were very helpful :)
– Rodrigo Castañon
Dec 10 '18 at 11:06
Glad to be of help! @RodrigoCastañon
– Safder
Dec 10 '18 at 22:32
add a comment |
I'm just taking my first complex analysis but from my understanding you are correct in the fact that pole z= 0 is of second order.
Using the formula for higher order poles.
$$Res_{z rightarrow x} = frac{1}{(n-1)!} lim_{zrightarrow x} frac{d^{n-1}}{dz^{n-1}}(z-x)^n f(z)$$
Where x is the pole 0 and n is the order.
Using this formula:
$$Res_{z rightarrow 0} = frac{1}{(2-1)!} lim_{zrightarrow 0} frac{d^{2-1}}{dz^{2-1}}(z-0)^2 f(z)$$
$$Res_{z rightarrow 0} = frac{1}{(1)!} lim_{zrightarrow x} frac{d}{dz}(z-0)^2 frac{1}{z(e^z - 1)}$$
$$ lim_{zrightarrow 0} frac{d}{dz}(z)^2 frac{1}{z(e^z - 1)}$$
$$ lim_{zrightarrow 0} frac{d}{dz}frac{z }{(e^z - 1)}$$
$$ lim_{zrightarrow 0} frac{(e^z - 1) - (e^z)z}{(e^z - 1)^2}$$
Now just do a couple L'Hopitals and you should get:
Res at z=0:
$$= - frac{1}{2}$$
answered Dec 10 '18 at 2:38
Safder
15110
I'm just taking my first complex analysis but from my understanding you are correct in the fact that pole z= 0 is of second order.
Using the formula for higher order poles.
$$Res_{z rightarrow x} = frac{1}{(n-1)!} lim_{zrightarrow x} frac{d^{n-1}}{dz^{n-1}}(z-x)^n f(z)$$
Where x is the pole 0 and n is the order.
Using this formula:
$$Res_{z rightarrow 0} = frac{1}{(2-1)!} lim_{zrightarrow 0} frac{d^{2-1}}{dz^{2-1}}(z-0)^2 f(z)$$
$$Res_{z rightarrow 0} = frac{1}{(1)!} lim_{zrightarrow x} frac{d}{dz}(z-0)^2 frac{1}{z(e^z - 1)}$$
$$ lim_{zrightarrow 0} frac{d}{dz}(z)^2 frac{1}{z(e^z - 1)}$$
$$ lim_{zrightarrow 0} frac{d}{dz}frac{z }{(e^z - 1)}$$
$$ lim_{zrightarrow 0} frac{(e^z - 1) - (e^z)z}{(e^z - 1)^2}$$
Now just do a couple L'Hopitals and you should get:
Res at z=0:
$$= - frac{1}{2}$$
answered Dec 10 '18 at 2:38
Safder
15110
edited Dec 10 '18 at 3:04
answered Dec 10 '18 at 2:38
Safder
15110
answered Dec 10 '18 at 2:38
Safder
15110
answered Dec 10 '18 at 2:38
Safder
15110
15110
1
Thank you! I've just started to study the subject too and your thoughts on the question were very helpful :)
– Rodrigo Castañon
Dec 10 '18 at 11:06
Glad to be of help! @RodrigoCastañon
– Safder
Dec 10 '18 at 22:32
add a comment |
1
Thank you! I've just started to study the subject too and your thoughts on the question were very helpful :)
– Rodrigo Castañon
Dec 10 '18 at 11:06
Glad to be of help! @RodrigoCastañon
– Safder
Dec 10 '18 at 22:32
1
1
Thank you! I've just started to study the subject too and your thoughts on the question were very helpful :)
– Rodrigo Castañon
Dec 10 '18 at 11:06
Thank you! I've just started to study the subject too and your thoughts on the question were very helpful :)
– Rodrigo Castañon
Dec 10 '18 at 11:06
Glad to be of help! @RodrigoCastañon
– Safder
Dec 10 '18 at 22:32
Glad to be of help! @RodrigoCastañon
– Safder
Dec 10 '18 at 22:32
add a comment |
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"wich doesn't meets the criteria explained before" Hmmm... why do you think it does not?
– Did
Dec 10 '18 at 2:50
@Did well, because g(z) has a singularity when z=0, doensn´t it? That would make it a non-analytic function in this point from what I know, but I could be wrong of course
– Rodrigo Castañon
Dec 10 '18 at 10:37
$g(z)$ has a removable singularity at $z=0$, since $lim_{zto 0} g(z) = 1$. You can make it analytic by defining $$ g(z) = begin{cases} dfrac{z}{e^z-1}, & z ne 0 \ 1, & z = 0 end{cases} $$
– Dylan
Dec 10 '18 at 14:42
How is the post below that you chose to accept, answering the question?
– Did
Dec 10 '18 at 15:16
@Dylan yes, I realized that after a while, thank you!
– Rodrigo Castañon
Dec 10 '18 at 17:51