Laurent Series expansion about the point $z_0 = i$ of $frac{z}{z^2+1}$












1












$begingroup$


I am trying to construct the Laurent series expansion of $f(z) = frac{z}{z^2+1}$ about $z_0 = i$ in the region ${z in mathbb{C}: 0 < |z - i| < 2}$ but I am stuck.



We can re-write $f(z) = dfrac{z}{z^2+1} = dfrac{z}{(z-i)(z+i)}$



This allows us to see more easily that there are first order poles at $pm i$. We can use partial fractions to obtain:



$$frac{1}{(z+i)(z-i)} = frac{i}{2(z+i)} - frac{i}{2(z-i)}$$



We want to re-write everything in terms of $z-i$:



$$frac{1}{z+i} = frac{1}{2i + (z - i)} = frac{1}{2i}left(frac{1}{1+frac{1}{2i}(z-i)}right)$$



Using the geometric series this gives:



$$ = frac{-i}{2} left(sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right)$$
Then we have that:



$$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$



But the problem is, I still have a term in the form of $z$ rather than $z-i$. How do I proceed from here?






complex-analysis taylor-expansion laurent-series







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  • $begingroup$
    $f(i)$ is not defined
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 18:27






  • 1




    $begingroup$
    @TitoEliatron You’re correct but that’s the use of Laurent series howellkb.uah.edu/MathPhysicsText/Complex_Variables/Laurent.pdf
    $endgroup$
    – Jane Doe
    Dec 2 '18 at 18:33












  • $begingroup$
    Sorry, I read Taylor, not Laurent.
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 18:38












  • $begingroup$
    You can write $z=(z-i)+i$ and proceed
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 18:39
















1












$begingroup$


I am trying to construct the Laurent series expansion of $f(z) = frac{z}{z^2+1}$ about $z_0 = i$ in the region ${z in mathbb{C}: 0 < |z - i| < 2}$ but I am stuck.



We can re-write $f(z) = dfrac{z}{z^2+1} = dfrac{z}{(z-i)(z+i)}$



This allows us to see more easily that there are first order poles at $pm i$. We can use partial fractions to obtain:



$$frac{1}{(z+i)(z-i)} = frac{i}{2(z+i)} - frac{i}{2(z-i)}$$



We want to re-write everything in terms of $z-i$:



$$frac{1}{z+i} = frac{1}{2i + (z - i)} = frac{1}{2i}left(frac{1}{1+frac{1}{2i}(z-i)}right)$$



Using the geometric series this gives:



$$ = frac{-i}{2} left(sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right)$$
Then we have that:



$$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$



But the problem is, I still have a term in the form of $z$ rather than $z-i$. How do I proceed from here?






complex-analysis taylor-expansion laurent-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    $f(i)$ is not defined
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 18:27






  • 1




    $begingroup$
    @TitoEliatron You’re correct but that’s the use of Laurent series howellkb.uah.edu/MathPhysicsText/Complex_Variables/Laurent.pdf
    $endgroup$
    – Jane Doe
    Dec 2 '18 at 18:33












  • $begingroup$
    Sorry, I read Taylor, not Laurent.
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 18:38












  • $begingroup$
    You can write $z=(z-i)+i$ and proceed
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 18:39














1












1








1





$begingroup$


I am trying to construct the Laurent series expansion of $f(z) = frac{z}{z^2+1}$ about $z_0 = i$ in the region ${z in mathbb{C}: 0 < |z - i| < 2}$ but I am stuck.



We can re-write $f(z) = dfrac{z}{z^2+1} = dfrac{z}{(z-i)(z+i)}$



This allows us to see more easily that there are first order poles at $pm i$. We can use partial fractions to obtain:



$$frac{1}{(z+i)(z-i)} = frac{i}{2(z+i)} - frac{i}{2(z-i)}$$



We want to re-write everything in terms of $z-i$:



$$frac{1}{z+i} = frac{1}{2i + (z - i)} = frac{1}{2i}left(frac{1}{1+frac{1}{2i}(z-i)}right)$$



Using the geometric series this gives:



$$ = frac{-i}{2} left(sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right)$$
Then we have that:



$$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$



But the problem is, I still have a term in the form of $z$ rather than $z-i$. How do I proceed from here?






complex-analysis taylor-expansion laurent-series







share|cite|improve this question











$endgroup$




I am trying to construct the Laurent series expansion of $f(z) = frac{z}{z^2+1}$ about $z_0 = i$ in the region ${z in mathbb{C}: 0 < |z - i| < 2}$ but I am stuck.



We can re-write $f(z) = dfrac{z}{z^2+1} = dfrac{z}{(z-i)(z+i)}$



This allows us to see more easily that there are first order poles at $pm i$. We can use partial fractions to obtain:



$$frac{1}{(z+i)(z-i)} = frac{i}{2(z+i)} - frac{i}{2(z-i)}$$



We want to re-write everything in terms of $z-i$:



$$frac{1}{z+i} = frac{1}{2i + (z - i)} = frac{1}{2i}left(frac{1}{1+frac{1}{2i}(z-i)}right)$$



Using the geometric series this gives:



$$ = frac{-i}{2} left(sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right)$$
Then we have that:



$$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$



But the problem is, I still have a term in the form of $z$ rather than $z-i$. How do I proceed from here?






complex-analysis taylor-expansion laurent-series




complex-analysis taylor-expansion laurent-series






share|cite|improve this question















share|cite|improve this question













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edited Dec 2 '18 at 19:51









Lorenzo B.

1,8402520




1,8402520










asked Dec 2 '18 at 18:21









Jane DoeJane Doe

19112




19112












  • $begingroup$
    $f(i)$ is not defined
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 18:27






  • 1




    $begingroup$
    @TitoEliatron You’re correct but that’s the use of Laurent series howellkb.uah.edu/MathPhysicsText/Complex_Variables/Laurent.pdf
    $endgroup$
    – Jane Doe
    Dec 2 '18 at 18:33












  • $begingroup$
    Sorry, I read Taylor, not Laurent.
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 18:38












  • $begingroup$
    You can write $z=(z-i)+i$ and proceed
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 18:39


















  • $begingroup$
    $f(i)$ is not defined
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 18:27






  • 1




    $begingroup$
    @TitoEliatron You’re correct but that’s the use of Laurent series howellkb.uah.edu/MathPhysicsText/Complex_Variables/Laurent.pdf
    $endgroup$
    – Jane Doe
    Dec 2 '18 at 18:33












  • $begingroup$
    Sorry, I read Taylor, not Laurent.
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 18:38












  • $begingroup$
    You can write $z=(z-i)+i$ and proceed
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 18:39
















$begingroup$
$f(i)$ is not defined
$endgroup$
– Tito Eliatron
Dec 2 '18 at 18:27




$begingroup$
$f(i)$ is not defined
$endgroup$
– Tito Eliatron
Dec 2 '18 at 18:27




1




1




$begingroup$
@TitoEliatron You’re correct but that’s the use of Laurent series howellkb.uah.edu/MathPhysicsText/Complex_Variables/Laurent.pdf
$endgroup$
– Jane Doe
Dec 2 '18 at 18:33






$begingroup$
@TitoEliatron You’re correct but that’s the use of Laurent series howellkb.uah.edu/MathPhysicsText/Complex_Variables/Laurent.pdf
$endgroup$
– Jane Doe
Dec 2 '18 at 18:33














$begingroup$
Sorry, I read Taylor, not Laurent.
$endgroup$
– Tito Eliatron
Dec 2 '18 at 18:38






$begingroup$
Sorry, I read Taylor, not Laurent.
$endgroup$
– Tito Eliatron
Dec 2 '18 at 18:38














$begingroup$
You can write $z=(z-i)+i$ and proceed
$endgroup$
– Tito Eliatron
Dec 2 '18 at 18:39




$begingroup$
You can write $z=(z-i)+i$ and proceed
$endgroup$
– Tito Eliatron
Dec 2 '18 at 18:39










2 Answers
2






active

oldest

votes


















1












$begingroup$

begin{align}&f(z)=frac{z}{z^2+1}=frac{A}{z-i}+frac{B}{z+i}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{z+i}end{align}
begin{align}&A=Res[f,i]=limlimits_{zrightarrow i}frac{z}{z+i}=frac{1}{2}
\&B=Res[f,-i]=limlimits_{zrightarrow -i}frac{z}{z-i}=frac{1}{2}end{align}begin{align}f(z)&= frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{2ileft(1+frac{z-i}{2i}right)}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}sum_{n=0}^{infty}frac{(-1)^n (z-i)^n}{(2i)^{n+1}}end{align}begin{align}=&frac{1}{2}frac{1}{z-i}-frac{i}{4}+frac{1}{8}(z-i)\&+frac{i}{16}(z-i)^2+cdotsend{align}
in the region ${zinmathbb{C}:0<|z-i|<2}.$






share|cite|improve this answer









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  • $begingroup$
    Thank you, that way makes much more sense!
    $endgroup$
    – Jane Doe
    Dec 3 '18 at 15:27



















0












$begingroup$

Following your argument,
$$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right) = (z-i+i)left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)\ = left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^{n+1}}{(2i)^n}right) - 1right) -left(frac{1}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$






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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

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    active

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    1












    $begingroup$

    begin{align}&f(z)=frac{z}{z^2+1}=frac{A}{z-i}+frac{B}{z+i}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{z+i}end{align}
    begin{align}&A=Res[f,i]=limlimits_{zrightarrow i}frac{z}{z+i}=frac{1}{2}
    \&B=Res[f,-i]=limlimits_{zrightarrow -i}frac{z}{z-i}=frac{1}{2}end{align}    begin{align}f(z)&= frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{2ileft(1+frac{z-i}{2i}right)}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}sum_{n=0}^{infty}frac{(-1)^n (z-i)^n}{(2i)^{n+1}}end{align}begin{align}=&frac{1}{2}frac{1}{z-i}-frac{i}{4}+frac{1}{8}(z-i)\&+frac{i}{16}(z-i)^2+cdotsend{align}
    in the region ${zinmathbb{C}:0<|z-i|<2}.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, that way makes much more sense!
      $endgroup$
      – Jane Doe
      Dec 3 '18 at 15:27
















    1












    $begingroup$

    begin{align}&f(z)=frac{z}{z^2+1}=frac{A}{z-i}+frac{B}{z+i}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{z+i}end{align}
    begin{align}&A=Res[f,i]=limlimits_{zrightarrow i}frac{z}{z+i}=frac{1}{2}
    \&B=Res[f,-i]=limlimits_{zrightarrow -i}frac{z}{z-i}=frac{1}{2}end{align}    begin{align}f(z)&= frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{2ileft(1+frac{z-i}{2i}right)}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}sum_{n=0}^{infty}frac{(-1)^n (z-i)^n}{(2i)^{n+1}}end{align}begin{align}=&frac{1}{2}frac{1}{z-i}-frac{i}{4}+frac{1}{8}(z-i)\&+frac{i}{16}(z-i)^2+cdotsend{align}
    in the region ${zinmathbb{C}:0<|z-i|<2}.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, that way makes much more sense!
      $endgroup$
      – Jane Doe
      Dec 3 '18 at 15:27














    1












    1








    1





    $begingroup$

    begin{align}&f(z)=frac{z}{z^2+1}=frac{A}{z-i}+frac{B}{z+i}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{z+i}end{align}
    begin{align}&A=Res[f,i]=limlimits_{zrightarrow i}frac{z}{z+i}=frac{1}{2}
    \&B=Res[f,-i]=limlimits_{zrightarrow -i}frac{z}{z-i}=frac{1}{2}end{align}    begin{align}f(z)&= frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{2ileft(1+frac{z-i}{2i}right)}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}sum_{n=0}^{infty}frac{(-1)^n (z-i)^n}{(2i)^{n+1}}end{align}begin{align}=&frac{1}{2}frac{1}{z-i}-frac{i}{4}+frac{1}{8}(z-i)\&+frac{i}{16}(z-i)^2+cdotsend{align}
    in the region ${zinmathbb{C}:0<|z-i|<2}.$






    share|cite|improve this answer









    $endgroup$



    begin{align}&f(z)=frac{z}{z^2+1}=frac{A}{z-i}+frac{B}{z+i}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{z+i}end{align}
    begin{align}&A=Res[f,i]=limlimits_{zrightarrow i}frac{z}{z+i}=frac{1}{2}
    \&B=Res[f,-i]=limlimits_{zrightarrow -i}frac{z}{z-i}=frac{1}{2}end{align}    begin{align}f(z)&= frac{1}{2}frac{1}{z-i}+frac{1}{2}frac{1}{2ileft(1+frac{z-i}{2i}right)}\&=frac{1}{2}frac{1}{z-i}+frac{1}{2}sum_{n=0}^{infty}frac{(-1)^n (z-i)^n}{(2i)^{n+1}}end{align}begin{align}=&frac{1}{2}frac{1}{z-i}-frac{i}{4}+frac{1}{8}(z-i)\&+frac{i}{16}(z-i)^2+cdotsend{align}
    in the region ${zinmathbb{C}:0<|z-i|<2}.$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 2 '18 at 22:19







    user621367



















    • $begingroup$
      Thank you, that way makes much more sense!
      $endgroup$
      – Jane Doe
      Dec 3 '18 at 15:27


















    • $begingroup$
      Thank you, that way makes much more sense!
      $endgroup$
      – Jane Doe
      Dec 3 '18 at 15:27
















    $begingroup$
    Thank you, that way makes much more sense!
    $endgroup$
    – Jane Doe
    Dec 3 '18 at 15:27




    $begingroup$
    Thank you, that way makes much more sense!
    $endgroup$
    – Jane Doe
    Dec 3 '18 at 15:27











    0












    $begingroup$

    Following your argument,
    $$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right) = (z-i+i)left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)\ = left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^{n+1}}{(2i)^n}right) - 1right) -left(frac{1}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Following your argument,
      $$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right) = (z-i+i)left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)\ = left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^{n+1}}{(2i)^n}right) - 1right) -left(frac{1}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Following your argument,
        $$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right) = (z-i+i)left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)\ = left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^{n+1}}{(2i)^n}right) - 1right) -left(frac{1}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$






        share|cite|improve this answer









        $endgroup$



        Following your argument,
        $$f(z) = zleft(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right) = (z-i+i)left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)\ = left(frac{i}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^{n+1}}{(2i)^n}right) - 1right) -left(frac{1}{2}right)left(left(frac{-i}{2}sum_{n=0}^{infty} (-1)^n frac{(z-i)^n}{(2i)^n}right) - frac{1}{z-i}right)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 18:42









        Tito EliatronTito Eliatron

        1,448622




        1,448622






























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            Le Mesnil-Réaume

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            Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
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            Ida-Boy-Ed-Garten

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            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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            web3.py web3.isConnected() returns false always

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            1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
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