Laurent Expansion $frace^zz-1$

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Expand $frace^zz-1$ at $|z-1|>1$



$t=z-1iff t+1=z$



$e^z=sum_n=0^infty fracz^nn!=sum_n=0^infty frac(t+1)^nn!$



$frac1t=frac11-1+t$



Is it the way?




complex-analysis laurent-series




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  • 1




    Note that $frac1z-1$ is already on the desired form, i.e. the Laurent series of $frac1z-1$ about $z=1$ is just $frac1z-1$.
    – Winther
    Jul 16 at 14:38











  • @Winther So I just need to multiply it by the Laurent series of $e^z-1$?
    – newhere
    Jul 16 at 14:41






  • 1




    Yes that's all you need to do here.
    – Winther
    Jul 16 at 14:42










  • @Winther shouldn't it be of the form $sum a_n(z-1)^n$? why is it enough, thanks you so much
    – newhere
    Jul 16 at 14:44






  • 1




    Yes it should and that's exactly what you will find: $e^z-1/(z-1) = [1 + (z-1) + (z-1)^2/2+ldots] / (z-1) = (z-1)^-1 + 1cdot (z-1)^0 + (z-1)^1/2 + ldots$
    – Winther
    Jul 16 at 14:44















up vote
0
down vote

favorite












Expand $frace^zz-1$ at $|z-1|>1$



$t=z-1iff t+1=z$



$e^z=sum_n=0^infty fracz^nn!=sum_n=0^infty frac(t+1)^nn!$



$frac1t=frac11-1+t$



Is it the way?




complex-analysis laurent-series




share|cite|improve this question















  • 1




    Note that $frac1z-1$ is already on the desired form, i.e. the Laurent series of $frac1z-1$ about $z=1$ is just $frac1z-1$.
    – Winther
    Jul 16 at 14:38











  • @Winther So I just need to multiply it by the Laurent series of $e^z-1$?
    – newhere
    Jul 16 at 14:41






  • 1




    Yes that's all you need to do here.
    – Winther
    Jul 16 at 14:42










  • @Winther shouldn't it be of the form $sum a_n(z-1)^n$? why is it enough, thanks you so much
    – newhere
    Jul 16 at 14:44






  • 1




    Yes it should and that's exactly what you will find: $e^z-1/(z-1) = [1 + (z-1) + (z-1)^2/2+ldots] / (z-1) = (z-1)^-1 + 1cdot (z-1)^0 + (z-1)^1/2 + ldots$
    – Winther
    Jul 16 at 14:44













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Expand $frace^zz-1$ at $|z-1|>1$



$t=z-1iff t+1=z$



$e^z=sum_n=0^infty fracz^nn!=sum_n=0^infty frac(t+1)^nn!$



$frac1t=frac11-1+t$



Is it the way?




complex-analysis laurent-series




share|cite|improve this question











Expand $frace^zz-1$ at $|z-1|>1$



$t=z-1iff t+1=z$



$e^z=sum_n=0^infty fracz^nn!=sum_n=0^infty frac(t+1)^nn!$



$frac1t=frac11-1+t$



Is it the way?





complex-analysis laurent-series





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 16 at 14:19









newhere

759310




759310







  • 1




    Note that $frac1z-1$ is already on the desired form, i.e. the Laurent series of $frac1z-1$ about $z=1$ is just $frac1z-1$.
    – Winther
    Jul 16 at 14:38











  • @Winther So I just need to multiply it by the Laurent series of $e^z-1$?
    – newhere
    Jul 16 at 14:41






  • 1




    Yes that's all you need to do here.
    – Winther
    Jul 16 at 14:42










  • @Winther shouldn't it be of the form $sum a_n(z-1)^n$? why is it enough, thanks you so much
    – newhere
    Jul 16 at 14:44






  • 1




    Yes it should and that's exactly what you will find: $e^z-1/(z-1) = [1 + (z-1) + (z-1)^2/2+ldots] / (z-1) = (z-1)^-1 + 1cdot (z-1)^0 + (z-1)^1/2 + ldots$
    – Winther
    Jul 16 at 14:44













  • 1




    Note that $frac1z-1$ is already on the desired form, i.e. the Laurent series of $frac1z-1$ about $z=1$ is just $frac1z-1$.
    – Winther
    Jul 16 at 14:38











  • @Winther So I just need to multiply it by the Laurent series of $e^z-1$?
    – newhere
    Jul 16 at 14:41






  • 1




    Yes that's all you need to do here.
    – Winther
    Jul 16 at 14:42










  • @Winther shouldn't it be of the form $sum a_n(z-1)^n$? why is it enough, thanks you so much
    – newhere
    Jul 16 at 14:44






  • 1




    Yes it should and that's exactly what you will find: $e^z-1/(z-1) = [1 + (z-1) + (z-1)^2/2+ldots] / (z-1) = (z-1)^-1 + 1cdot (z-1)^0 + (z-1)^1/2 + ldots$
    – Winther
    Jul 16 at 14:44








1




1




Note that $frac1z-1$ is already on the desired form, i.e. the Laurent series of $frac1z-1$ about $z=1$ is just $frac1z-1$.
– Winther
Jul 16 at 14:38





Note that $frac1z-1$ is already on the desired form, i.e. the Laurent series of $frac1z-1$ about $z=1$ is just $frac1z-1$.
– Winther
Jul 16 at 14:38













@Winther So I just need to multiply it by the Laurent series of $e^z-1$?
– newhere
Jul 16 at 14:41




@Winther So I just need to multiply it by the Laurent series of $e^z-1$?
– newhere
Jul 16 at 14:41




1




1




Yes that's all you need to do here.
– Winther
Jul 16 at 14:42




Yes that's all you need to do here.
– Winther
Jul 16 at 14:42












@Winther shouldn't it be of the form $sum a_n(z-1)^n$? why is it enough, thanks you so much
– newhere
Jul 16 at 14:44




@Winther shouldn't it be of the form $sum a_n(z-1)^n$? why is it enough, thanks you so much
– newhere
Jul 16 at 14:44




1




1




Yes it should and that's exactly what you will find: $e^z-1/(z-1) = [1 + (z-1) + (z-1)^2/2+ldots] / (z-1) = (z-1)^-1 + 1cdot (z-1)^0 + (z-1)^1/2 + ldots$
– Winther
Jul 16 at 14:44





Yes it should and that's exactly what you will find: $e^z-1/(z-1) = [1 + (z-1) + (z-1)^2/2+ldots] / (z-1) = (z-1)^-1 + 1cdot (z-1)^0 + (z-1)^1/2 + ldots$
– Winther
Jul 16 at 14:44











1 Answer
1






active

oldest

votes

















up vote
3
down vote



accepted










Simpler:



Don't forget the radius of convergence of the exponential function if infinite), so you can write



$$rm e^z=rm e, e^z-1=rm eBigl(1+(z-1)+frac(z-1)^22+dots+frac(z-1)^nn!+dotsmBigr).$$






share|cite|improve this answer





















  • what I can I do with $frac1z-1$ to add $z-1+1-1$?
    – newhere
    Jul 16 at 14:36






  • 1




    I don't understand what you want to add. You only have to divide the above expansion by $z-1$. The Laurent expansion here starts at degree $-1$.
    – Bernard
    Jul 16 at 14:54










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










Simpler:



Don't forget the radius of convergence of the exponential function if infinite), so you can write



$$rm e^z=rm e, e^z-1=rm eBigl(1+(z-1)+frac(z-1)^22+dots+frac(z-1)^nn!+dotsmBigr).$$






share|cite|improve this answer





















  • what I can I do with $frac1z-1$ to add $z-1+1-1$?
    – newhere
    Jul 16 at 14:36






  • 1




    I don't understand what you want to add. You only have to divide the above expansion by $z-1$. The Laurent expansion here starts at degree $-1$.
    – Bernard
    Jul 16 at 14:54














up vote
3
down vote



accepted










Simpler:



Don't forget the radius of convergence of the exponential function if infinite), so you can write



$$rm e^z=rm e, e^z-1=rm eBigl(1+(z-1)+frac(z-1)^22+dots+frac(z-1)^nn!+dotsmBigr).$$






share|cite|improve this answer





















  • what I can I do with $frac1z-1$ to add $z-1+1-1$?
    – newhere
    Jul 16 at 14:36






  • 1




    I don't understand what you want to add. You only have to divide the above expansion by $z-1$. The Laurent expansion here starts at degree $-1$.
    – Bernard
    Jul 16 at 14:54












up vote
3
down vote



accepted







up vote
3
down vote



accepted






Simpler:



Don't forget the radius of convergence of the exponential function if infinite), so you can write



$$rm e^z=rm e, e^z-1=rm eBigl(1+(z-1)+frac(z-1)^22+dots+frac(z-1)^nn!+dotsmBigr).$$






share|cite|improve this answer













Simpler:



Don't forget the radius of convergence of the exponential function if infinite), so you can write



$$rm e^z=rm e, e^z-1=rm eBigl(1+(z-1)+frac(z-1)^22+dots+frac(z-1)^nn!+dotsmBigr).$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 16 at 14:33









Bernard

110k635103




110k635103











  • what I can I do with $frac1z-1$ to add $z-1+1-1$?
    – newhere
    Jul 16 at 14:36






  • 1




    I don't understand what you want to add. You only have to divide the above expansion by $z-1$. The Laurent expansion here starts at degree $-1$.
    – Bernard
    Jul 16 at 14:54
















  • what I can I do with $frac1z-1$ to add $z-1+1-1$?
    – newhere
    Jul 16 at 14:36






  • 1




    I don't understand what you want to add. You only have to divide the above expansion by $z-1$. The Laurent expansion here starts at degree $-1$.
    – Bernard
    Jul 16 at 14:54















what I can I do with $frac1z-1$ to add $z-1+1-1$?
– newhere
Jul 16 at 14:36




what I can I do with $frac1z-1$ to add $z-1+1-1$?
– newhere
Jul 16 at 14:36




1




1




I don't understand what you want to add. You only have to divide the above expansion by $z-1$. The Laurent expansion here starts at degree $-1$.
– Bernard
Jul 16 at 14:54




I don't understand what you want to add. You only have to divide the above expansion by $z-1$. The Laurent expansion here starts at degree $-1$.
– Bernard
Jul 16 at 14:54












 

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