Expanding Laurent Series with $(z-1)$ term on numerator












1












$begingroup$


$frac{z-1}{z^3cdot(z-2)}$. I want to expand series in regions $|z|<2$. I tried to just pull out $frac{(z-1)}{z^3}$ and expand the $frac{1}{(z-2)}$,the answer I got is similar to solution but the solution doesn't have the $(z-1)$ part. What do I do with that term? I've tried making it $frac{1}{(z-1)^{-1}}$ and take partial fractions, do it from there but still wrong.






complex-analysis laurent-series







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$endgroup$

















    1












    $begingroup$


    $frac{z-1}{z^3cdot(z-2)}$. I want to expand series in regions $|z|<2$. I tried to just pull out $frac{(z-1)}{z^3}$ and expand the $frac{1}{(z-2)}$,the answer I got is similar to solution but the solution doesn't have the $(z-1)$ part. What do I do with that term? I've tried making it $frac{1}{(z-1)^{-1}}$ and take partial fractions, do it from there but still wrong.






    complex-analysis laurent-series







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      $frac{z-1}{z^3cdot(z-2)}$. I want to expand series in regions $|z|<2$. I tried to just pull out $frac{(z-1)}{z^3}$ and expand the $frac{1}{(z-2)}$,the answer I got is similar to solution but the solution doesn't have the $(z-1)$ part. What do I do with that term? I've tried making it $frac{1}{(z-1)^{-1}}$ and take partial fractions, do it from there but still wrong.






      complex-analysis laurent-series







      share|cite|improve this question









      $endgroup$




      $frac{z-1}{z^3cdot(z-2)}$. I want to expand series in regions $|z|<2$. I tried to just pull out $frac{(z-1)}{z^3}$ and expand the $frac{1}{(z-2)}$,the answer I got is similar to solution but the solution doesn't have the $(z-1)$ part. What do I do with that term? I've tried making it $frac{1}{(z-1)^{-1}}$ and take partial fractions, do it from there but still wrong.






      complex-analysis laurent-series




      complex-analysis laurent-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 22 at 5:22









      MinYoung KimMinYoung Kim

      907




      907






















          1 Answer
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          $begingroup$

          First step: find $A,B,C$ and $D$ such that



          $$frac{z-1}{z^3cdot(z-2)}= frac{A}{z}+frac{B}{z^2}+frac{C}{z^3}+frac{D}{z-2}.$$



          Second step:



          $$frac{D}{z-2}=frac{D}{2}frac{1}{frac{z}{2}-1}=-frac{D}{2}sum_{n=0}^{infty}frac{z^n}{2^n}.$$



          Then the Laurent expansion is given by



          $$frac{z-1}{z^3cdot(z-2)}= frac{A}{z}+frac{B}{z^2}+frac{C}{z^3}-frac{D}{2}sum_{n=0}^{infty}frac{z^n}{2^n}$$



          for $0<|z|<2.$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Thanks. I forgot about how to do partial fraction on $frac{1}{z^3}$ repeated root. Also it is definitely easier to use geometric series expansion than binomial expansion(which i've been doing).
            $endgroup$
            – MinYoung Kim
            Jan 22 at 6:00













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          active

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          1












          $begingroup$

          First step: find $A,B,C$ and $D$ such that



          $$frac{z-1}{z^3cdot(z-2)}= frac{A}{z}+frac{B}{z^2}+frac{C}{z^3}+frac{D}{z-2}.$$



          Second step:



          $$frac{D}{z-2}=frac{D}{2}frac{1}{frac{z}{2}-1}=-frac{D}{2}sum_{n=0}^{infty}frac{z^n}{2^n}.$$



          Then the Laurent expansion is given by



          $$frac{z-1}{z^3cdot(z-2)}= frac{A}{z}+frac{B}{z^2}+frac{C}{z^3}-frac{D}{2}sum_{n=0}^{infty}frac{z^n}{2^n}$$



          for $0<|z|<2.$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Thanks. I forgot about how to do partial fraction on $frac{1}{z^3}$ repeated root. Also it is definitely easier to use geometric series expansion than binomial expansion(which i've been doing).
            $endgroup$
            – MinYoung Kim
            Jan 22 at 6:00


















          1












          $begingroup$

          First step: find $A,B,C$ and $D$ such that



          $$frac{z-1}{z^3cdot(z-2)}= frac{A}{z}+frac{B}{z^2}+frac{C}{z^3}+frac{D}{z-2}.$$



          Second step:



          $$frac{D}{z-2}=frac{D}{2}frac{1}{frac{z}{2}-1}=-frac{D}{2}sum_{n=0}^{infty}frac{z^n}{2^n}.$$



          Then the Laurent expansion is given by



          $$frac{z-1}{z^3cdot(z-2)}= frac{A}{z}+frac{B}{z^2}+frac{C}{z^3}-frac{D}{2}sum_{n=0}^{infty}frac{z^n}{2^n}$$



          for $0<|z|<2.$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Thanks. I forgot about how to do partial fraction on $frac{1}{z^3}$ repeated root. Also it is definitely easier to use geometric series expansion than binomial expansion(which i've been doing).
            $endgroup$
            – MinYoung Kim
            Jan 22 at 6:00
















          1












          1








          1





          $begingroup$

          First step: find $A,B,C$ and $D$ such that



          $$frac{z-1}{z^3cdot(z-2)}= frac{A}{z}+frac{B}{z^2}+frac{C}{z^3}+frac{D}{z-2}.$$



          Second step:



          $$frac{D}{z-2}=frac{D}{2}frac{1}{frac{z}{2}-1}=-frac{D}{2}sum_{n=0}^{infty}frac{z^n}{2^n}.$$



          Then the Laurent expansion is given by



          $$frac{z-1}{z^3cdot(z-2)}= frac{A}{z}+frac{B}{z^2}+frac{C}{z^3}-frac{D}{2}sum_{n=0}^{infty}frac{z^n}{2^n}$$



          for $0<|z|<2.$






          share|cite|improve this answer









          $endgroup$



          First step: find $A,B,C$ and $D$ such that



          $$frac{z-1}{z^3cdot(z-2)}= frac{A}{z}+frac{B}{z^2}+frac{C}{z^3}+frac{D}{z-2}.$$



          Second step:



          $$frac{D}{z-2}=frac{D}{2}frac{1}{frac{z}{2}-1}=-frac{D}{2}sum_{n=0}^{infty}frac{z^n}{2^n}.$$



          Then the Laurent expansion is given by



          $$frac{z-1}{z^3cdot(z-2)}= frac{A}{z}+frac{B}{z^2}+frac{C}{z^3}-frac{D}{2}sum_{n=0}^{infty}frac{z^n}{2^n}$$



          for $0<|z|<2.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 22 at 5:41









          FredFred

          47.8k1849




          47.8k1849








          • 1




            $begingroup$
            Thanks. I forgot about how to do partial fraction on $frac{1}{z^3}$ repeated root. Also it is definitely easier to use geometric series expansion than binomial expansion(which i've been doing).
            $endgroup$
            – MinYoung Kim
            Jan 22 at 6:00
















          • 1




            $begingroup$
            Thanks. I forgot about how to do partial fraction on $frac{1}{z^3}$ repeated root. Also it is definitely easier to use geometric series expansion than binomial expansion(which i've been doing).
            $endgroup$
            – MinYoung Kim
            Jan 22 at 6:00










          1




          1




          $begingroup$
          Thanks. I forgot about how to do partial fraction on $frac{1}{z^3}$ repeated root. Also it is definitely easier to use geometric series expansion than binomial expansion(which i've been doing).
          $endgroup$
          – MinYoung Kim
          Jan 22 at 6:00






          $begingroup$
          Thanks. I forgot about how to do partial fraction on $frac{1}{z^3}$ repeated root. Also it is definitely easier to use geometric series expansion than binomial expansion(which i've been doing).
          $endgroup$
          – MinYoung Kim
          Jan 22 at 6:00




















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