Finding Laurent series of a function by changing variable

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I'm trying to find Laurent series of the following function at $$mid zmid<1$$
My function is as follows:
$$f(z)=dfrac1z(z-1)(z-2)=dfrac1zleft(dfrac1(z-2)+dfrac1(1-z)right)$$



My attempt was like this :
$$dfrac1zleft(dfrac-11+(1-z)+sum_k=0^nz^kright)=dfrac1zleft(dfrac-11-omega+sum_k=0^nz^kright)$$
So by calling $$omega=z-1$$ I made it Taylor series expandable. But I found the wrong answer. Is it because by changing variable I expended my function for point 1 ?




complex-analysis taylor-expansion laurent-series




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  • 1




    You are trying to find the LS for $|z|<1$. Expand around $z=0$.
    – Mark Viola
    Jul 29 at 19:53










  • and when I substitute omega variable I will find taylor series of that term for point z=1 right? That is where I'm wrong
    – GGphys
    Jul 29 at 19:55














up vote
1
down vote

favorite












I'm trying to find Laurent series of the following function at $$mid zmid<1$$
My function is as follows:
$$f(z)=dfrac1z(z-1)(z-2)=dfrac1zleft(dfrac1(z-2)+dfrac1(1-z)right)$$



My attempt was like this :
$$dfrac1zleft(dfrac-11+(1-z)+sum_k=0^nz^kright)=dfrac1zleft(dfrac-11-omega+sum_k=0^nz^kright)$$
So by calling $$omega=z-1$$ I made it Taylor series expandable. But I found the wrong answer. Is it because by changing variable I expended my function for point 1 ?




complex-analysis taylor-expansion laurent-series




share|cite|improve this question

















  • 1




    You are trying to find the LS for $|z|<1$. Expand around $z=0$.
    – Mark Viola
    Jul 29 at 19:53










  • and when I substitute omega variable I will find taylor series of that term for point z=1 right? That is where I'm wrong
    – GGphys
    Jul 29 at 19:55












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I'm trying to find Laurent series of the following function at $$mid zmid<1$$
My function is as follows:
$$f(z)=dfrac1z(z-1)(z-2)=dfrac1zleft(dfrac1(z-2)+dfrac1(1-z)right)$$



My attempt was like this :
$$dfrac1zleft(dfrac-11+(1-z)+sum_k=0^nz^kright)=dfrac1zleft(dfrac-11-omega+sum_k=0^nz^kright)$$
So by calling $$omega=z-1$$ I made it Taylor series expandable. But I found the wrong answer. Is it because by changing variable I expended my function for point 1 ?




complex-analysis taylor-expansion laurent-series




share|cite|improve this question













I'm trying to find Laurent series of the following function at $$mid zmid<1$$
My function is as follows:
$$f(z)=dfrac1z(z-1)(z-2)=dfrac1zleft(dfrac1(z-2)+dfrac1(1-z)right)$$



My attempt was like this :
$$dfrac1zleft(dfrac-11+(1-z)+sum_k=0^nz^kright)=dfrac1zleft(dfrac-11-omega+sum_k=0^nz^kright)$$
So by calling $$omega=z-1$$ I made it Taylor series expandable. But I found the wrong answer. Is it because by changing variable I expended my function for point 1 ?





complex-analysis taylor-expansion laurent-series





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 29 at 20:05
























asked Jul 29 at 19:51









GGphys

876




876







  • 1




    You are trying to find the LS for $|z|<1$. Expand around $z=0$.
    – Mark Viola
    Jul 29 at 19:53










  • and when I substitute omega variable I will find taylor series of that term for point z=1 right? That is where I'm wrong
    – GGphys
    Jul 29 at 19:55












  • 1




    You are trying to find the LS for $|z|<1$. Expand around $z=0$.
    – Mark Viola
    Jul 29 at 19:53










  • and when I substitute omega variable I will find taylor series of that term for point z=1 right? That is where I'm wrong
    – GGphys
    Jul 29 at 19:55







1




1




You are trying to find the LS for $|z|<1$. Expand around $z=0$.
– Mark Viola
Jul 29 at 19:53




You are trying to find the LS for $|z|<1$. Expand around $z=0$.
– Mark Viola
Jul 29 at 19:53












and when I substitute omega variable I will find taylor series of that term for point z=1 right? That is where I'm wrong
– GGphys
Jul 29 at 19:55




and when I substitute omega variable I will find taylor series of that term for point z=1 right? That is where I'm wrong
– GGphys
Jul 29 at 19:55










1 Answer
1






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1
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accepted










HINTS:



Note that for $|z|<1$,



$$frac11-z=sum_n=0^infty z^n$$



and for $|z|<2<1$



$$frac12-z=frac12sum_n=0^infty left(fracz2right)^n$$






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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    HINTS:



    Note that for $|z|<1$,



    $$frac11-z=sum_n=0^infty z^n$$



    and for $|z|<2<1$



    $$frac12-z=frac12sum_n=0^infty left(fracz2right)^n$$






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      HINTS:



      Note that for $|z|<1$,



      $$frac11-z=sum_n=0^infty z^n$$



      and for $|z|<2<1$



      $$frac12-z=frac12sum_n=0^infty left(fracz2right)^n$$






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        HINTS:



        Note that for $|z|<1$,



        $$frac11-z=sum_n=0^infty z^n$$



        and for $|z|<2<1$



        $$frac12-z=frac12sum_n=0^infty left(fracz2right)^n$$






        share|cite|improve this answer













        HINTS:



        Note that for $|z|<1$,



        $$frac11-z=sum_n=0^infty z^n$$



        and for $|z|<2<1$



        $$frac12-z=frac12sum_n=0^infty left(fracz2right)^n$$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 29 at 19:55









        Mark Viola

        126k1172167




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