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

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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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            Bundesstraße 106

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            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
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            Verónica Boquete

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            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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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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