Laurent series of $ frac{z-12}{z^2 + z - 6}$ for $|z-1|>4$











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How do you find the Laurent series for $f(z) = dfrac{z-12}{z^2 + z - 6}$ valid for $|z-1|>4$?




I know that $f(z) = dfrac{z-12}{z^2 + z - 6} = dfrac{-2}{z-2} + dfrac{3}{z+3}$



It is easy for me to extract a series for $dfrac{3}{z+3}$, but have no idea how to do it for $dfrac{-2}{z-2}$.



Please help? Thank you!






complex-analysis laurent-series







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    BTW, $f$ is not a polynomial.
    – lhf
    yesterday

















up vote
1
down vote

favorite













How do you find the Laurent series for $f(z) = dfrac{z-12}{z^2 + z - 6}$ valid for $|z-1|>4$?




I know that $f(z) = dfrac{z-12}{z^2 + z - 6} = dfrac{-2}{z-2} + dfrac{3}{z+3}$



It is easy for me to extract a series for $dfrac{3}{z+3}$, but have no idea how to do it for $dfrac{-2}{z-2}$.



Please help? Thank you!






complex-analysis laurent-series







share|cite|improve this question




















  • 2




    BTW, $f$ is not a polynomial.
    – lhf
    yesterday















up vote
1
down vote

favorite









up vote
1
down vote

favorite












How do you find the Laurent series for $f(z) = dfrac{z-12}{z^2 + z - 6}$ valid for $|z-1|>4$?




I know that $f(z) = dfrac{z-12}{z^2 + z - 6} = dfrac{-2}{z-2} + dfrac{3}{z+3}$



It is easy for me to extract a series for $dfrac{3}{z+3}$, but have no idea how to do it for $dfrac{-2}{z-2}$.



Please help? Thank you!






complex-analysis laurent-series







share|cite|improve this question
















How do you find the Laurent series for $f(z) = dfrac{z-12}{z^2 + z - 6}$ valid for $|z-1|>4$?




I know that $f(z) = dfrac{z-12}{z^2 + z - 6} = dfrac{-2}{z-2} + dfrac{3}{z+3}$



It is easy for me to extract a series for $dfrac{3}{z+3}$, but have no idea how to do it for $dfrac{-2}{z-2}$.



Please help? Thank you!






complex-analysis laurent-series




complex-analysis laurent-series






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edited yesterday









Robert Z

90k1056128




90k1056128










asked yesterday









Jonelle Yu

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1696








  • 2




    BTW, $f$ is not a polynomial.
    – lhf
    yesterday
















  • 2




    BTW, $f$ is not a polynomial.
    – lhf
    yesterday










2




2




BTW, $f$ is not a polynomial.
– lhf
yesterday






BTW, $f$ is not a polynomial.
– lhf
yesterday












2 Answers
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Hint. Starting from your partial fraction decomposition, we have that
$$frac{z-12}{z^2 + z - 6} = -frac{2}{u-1} + frac{3}{u+4}=-frac{2/u}{1-1/u} + frac{3/u}{1+4/u}$$
where $u=z-1$. Now note that $1/|u|<4/|u|<1$ when $|z-1|>4$.






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    Use the fact thatbegin{align}frac{-2}{z-2}&=frac{-2}{-1+(z-1)}\&=frac2{1-(z-1)}\&=-2sum_{n=-infty}^{-1}(z-1)^nend{align}and thatbegin{align}frac3{z+3}&=frac3{4+(z-1)}\&=frac34timesfrac1{1+frac{z-1}4}\&=-frac34sum_{n=-infty}^{-1}frac{(-1)^n}{4^n}(z-1)^n.end{align}






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      2 Answers
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      2 Answers
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      Hint. Starting from your partial fraction decomposition, we have that
      $$frac{z-12}{z^2 + z - 6} = -frac{2}{u-1} + frac{3}{u+4}=-frac{2/u}{1-1/u} + frac{3/u}{1+4/u}$$
      where $u=z-1$. Now note that $1/|u|<4/|u|<1$ when $|z-1|>4$.






      share|cite|improve this answer

























        up vote
        1
        down vote













        Hint. Starting from your partial fraction decomposition, we have that
        $$frac{z-12}{z^2 + z - 6} = -frac{2}{u-1} + frac{3}{u+4}=-frac{2/u}{1-1/u} + frac{3/u}{1+4/u}$$
        where $u=z-1$. Now note that $1/|u|<4/|u|<1$ when $|z-1|>4$.






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          Hint. Starting from your partial fraction decomposition, we have that
          $$frac{z-12}{z^2 + z - 6} = -frac{2}{u-1} + frac{3}{u+4}=-frac{2/u}{1-1/u} + frac{3/u}{1+4/u}$$
          where $u=z-1$. Now note that $1/|u|<4/|u|<1$ when $|z-1|>4$.






          share|cite|improve this answer












          Hint. Starting from your partial fraction decomposition, we have that
          $$frac{z-12}{z^2 + z - 6} = -frac{2}{u-1} + frac{3}{u+4}=-frac{2/u}{1-1/u} + frac{3/u}{1+4/u}$$
          where $u=z-1$. Now note that $1/|u|<4/|u|<1$ when $|z-1|>4$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Robert Z

          90k1056128




          90k1056128






















              up vote
              0
              down vote













              Use the fact thatbegin{align}frac{-2}{z-2}&=frac{-2}{-1+(z-1)}\&=frac2{1-(z-1)}\&=-2sum_{n=-infty}^{-1}(z-1)^nend{align}and thatbegin{align}frac3{z+3}&=frac3{4+(z-1)}\&=frac34timesfrac1{1+frac{z-1}4}\&=-frac34sum_{n=-infty}^{-1}frac{(-1)^n}{4^n}(z-1)^n.end{align}






              share|cite|improve this answer

























                up vote
                0
                down vote













                Use the fact thatbegin{align}frac{-2}{z-2}&=frac{-2}{-1+(z-1)}\&=frac2{1-(z-1)}\&=-2sum_{n=-infty}^{-1}(z-1)^nend{align}and thatbegin{align}frac3{z+3}&=frac3{4+(z-1)}\&=frac34timesfrac1{1+frac{z-1}4}\&=-frac34sum_{n=-infty}^{-1}frac{(-1)^n}{4^n}(z-1)^n.end{align}






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Use the fact thatbegin{align}frac{-2}{z-2}&=frac{-2}{-1+(z-1)}\&=frac2{1-(z-1)}\&=-2sum_{n=-infty}^{-1}(z-1)^nend{align}and thatbegin{align}frac3{z+3}&=frac3{4+(z-1)}\&=frac34timesfrac1{1+frac{z-1}4}\&=-frac34sum_{n=-infty}^{-1}frac{(-1)^n}{4^n}(z-1)^n.end{align}






                  share|cite|improve this answer












                  Use the fact thatbegin{align}frac{-2}{z-2}&=frac{-2}{-1+(z-1)}\&=frac2{1-(z-1)}\&=-2sum_{n=-infty}^{-1}(z-1)^nend{align}and thatbegin{align}frac3{z+3}&=frac3{4+(z-1)}\&=frac34timesfrac1{1+frac{z-1}4}\&=-frac34sum_{n=-infty}^{-1}frac{(-1)^n}{4^n}(z-1)^n.end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  José Carlos Santos

                  139k18111203




                  139k18111203






























                       

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