Finding laurent series expansion at Infinity












0















Find the laurent series expansion at $infty$ of the follwoing function:



$dfrac{1}{z^2-8z+25}$




Consider $dfrac{1}{z^2-8z+25}$,Replace $z=frac{1}{w}$,we get



$dfrac{1}{z^2-8z+25}=dfrac{w^2}{25(w^2-frac{8}{25}w+frac{1}{25})}$



$(w^2-frac{8}{25}w+frac{1}{25})=(w-frac{4}{25})^2+(frac{3}{25})^2$



Hence $dfrac{w^2}{25(w^2-frac{8}{25}w+frac{1}{25})}$



$=dfrac{w^2}{25((frac{3}{25})^2{1+dfrac{(w-frac{4}{25})^2}{frac{3}{25}}})} $



The next step involves just the expansion of ${1+dfrac{(w-frac{4}{25})^2}{frac{3}{25}})}^{-1}$



I dont understand if its going in the right direction or not



Can someone please help if I should proceed further






complex-analysis laurent-series







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    0















    Find the laurent series expansion at $infty$ of the follwoing function:



    $dfrac{1}{z^2-8z+25}$




    Consider $dfrac{1}{z^2-8z+25}$,Replace $z=frac{1}{w}$,we get



    $dfrac{1}{z^2-8z+25}=dfrac{w^2}{25(w^2-frac{8}{25}w+frac{1}{25})}$



    $(w^2-frac{8}{25}w+frac{1}{25})=(w-frac{4}{25})^2+(frac{3}{25})^2$



    Hence $dfrac{w^2}{25(w^2-frac{8}{25}w+frac{1}{25})}$



    $=dfrac{w^2}{25((frac{3}{25})^2{1+dfrac{(w-frac{4}{25})^2}{frac{3}{25}}})} $



    The next step involves just the expansion of ${1+dfrac{(w-frac{4}{25})^2}{frac{3}{25}})}^{-1}$



    I dont understand if its going in the right direction or not



    Can someone please help if I should proceed further






    complex-analysis laurent-series







    share|cite|improve this question

























      0












      0








      0








      Find the laurent series expansion at $infty$ of the follwoing function:



      $dfrac{1}{z^2-8z+25}$




      Consider $dfrac{1}{z^2-8z+25}$,Replace $z=frac{1}{w}$,we get



      $dfrac{1}{z^2-8z+25}=dfrac{w^2}{25(w^2-frac{8}{25}w+frac{1}{25})}$



      $(w^2-frac{8}{25}w+frac{1}{25})=(w-frac{4}{25})^2+(frac{3}{25})^2$



      Hence $dfrac{w^2}{25(w^2-frac{8}{25}w+frac{1}{25})}$



      $=dfrac{w^2}{25((frac{3}{25})^2{1+dfrac{(w-frac{4}{25})^2}{frac{3}{25}}})} $



      The next step involves just the expansion of ${1+dfrac{(w-frac{4}{25})^2}{frac{3}{25}})}^{-1}$



      I dont understand if its going in the right direction or not



      Can someone please help if I should proceed further






      complex-analysis laurent-series







      share|cite|improve this question














      Find the laurent series expansion at $infty$ of the follwoing function:



      $dfrac{1}{z^2-8z+25}$




      Consider $dfrac{1}{z^2-8z+25}$,Replace $z=frac{1}{w}$,we get



      $dfrac{1}{z^2-8z+25}=dfrac{w^2}{25(w^2-frac{8}{25}w+frac{1}{25})}$



      $(w^2-frac{8}{25}w+frac{1}{25})=(w-frac{4}{25})^2+(frac{3}{25})^2$



      Hence $dfrac{w^2}{25(w^2-frac{8}{25}w+frac{1}{25})}$



      $=dfrac{w^2}{25((frac{3}{25})^2{1+dfrac{(w-frac{4}{25})^2}{frac{3}{25}}})} $



      The next step involves just the expansion of ${1+dfrac{(w-frac{4}{25})^2}{frac{3}{25}})}^{-1}$



      I dont understand if its going in the right direction or not



      Can someone please help if I should proceed further






      complex-analysis laurent-series




      complex-analysis laurent-series






      share|cite|improve this question













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      asked Nov 28 '18 at 11:37









      Join_PhD

      3068




      3068






















          3 Answers
          3






          active

          oldest

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          2














          Hint:



          $dfrac{1}{z^2-8z+25}=dfrac{w^2}{1-(8w-25w^2)}$



          Now use



          $dfrac{1}{(1-a)}=1+a+a^2....$ formula above






          share|cite|improve this answer























          • But what is the problem with my steps
            – Join_PhD
            Nov 28 '18 at 11:54










          • In this manner, you can approach in less number of steps because you know expansion
            – Shubham
            Nov 28 '18 at 11:55










          • @Shubham: Edit $dfrac{1}{1-a}$.
            – Yadati Kiran
            Nov 28 '18 at 12:00










          • Thanks. Edited @YadatiKiran
            – Shubham
            Nov 28 '18 at 12:03



















          1














          Alternatively, notice that
          begin{align}
          frac{1}{z^2-8z+25} &= frac{1}{(z-4+3i)(z-4-3i)} = frac{1}{6i}left(frac{1}{z-4-3i}-frac{1}{z-4+3i}right)\
          &= frac{1}{6i}left(frac{w}{1-w(4+3i)}-frac{w}{1-w(4-3i)}right),
          end{align}
          and then use the geometric series formula on each of the terms.






          share|cite|improve this answer























          • the way what i have done is it correct?
            – Join_PhD
            Nov 28 '18 at 11:55










          • Your steps seem correct. They also don't seem to lead to anything fruitful.
            – MisterRiemann
            Nov 28 '18 at 11:56



















          1














          No, you are not in the right direction. You should perform a partial fractions decomposition:$$frac{w^2}{25left(w^2-frac8{25}w+frac1{25}right)}=frac{frac i6}{frac{4+3i}{25}-w}-frac{frac i6}{frac{4-3i}{25}-w}$$and then apply twice the equality $frac1{1-z}=1+z+z^2+z^2+cdots$






          share|cite|improve this answer





















          • But why is my direction not correct?
            – Join_PhD
            Nov 28 '18 at 11:59










          • I just don't see how you will get the correct answer from that.
            – José Carlos Santos
            Nov 28 '18 at 12:01













          Your Answer





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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          Hint:



          $dfrac{1}{z^2-8z+25}=dfrac{w^2}{1-(8w-25w^2)}$



          Now use



          $dfrac{1}{(1-a)}=1+a+a^2....$ formula above






          share|cite|improve this answer























          • But what is the problem with my steps
            – Join_PhD
            Nov 28 '18 at 11:54










          • In this manner, you can approach in less number of steps because you know expansion
            – Shubham
            Nov 28 '18 at 11:55










          • @Shubham: Edit $dfrac{1}{1-a}$.
            – Yadati Kiran
            Nov 28 '18 at 12:00










          • Thanks. Edited @YadatiKiran
            – Shubham
            Nov 28 '18 at 12:03
















          2














          Hint:



          $dfrac{1}{z^2-8z+25}=dfrac{w^2}{1-(8w-25w^2)}$



          Now use



          $dfrac{1}{(1-a)}=1+a+a^2....$ formula above






          share|cite|improve this answer























          • But what is the problem with my steps
            – Join_PhD
            Nov 28 '18 at 11:54










          • In this manner, you can approach in less number of steps because you know expansion
            – Shubham
            Nov 28 '18 at 11:55










          • @Shubham: Edit $dfrac{1}{1-a}$.
            – Yadati Kiran
            Nov 28 '18 at 12:00










          • Thanks. Edited @YadatiKiran
            – Shubham
            Nov 28 '18 at 12:03














          2












          2








          2






          Hint:



          $dfrac{1}{z^2-8z+25}=dfrac{w^2}{1-(8w-25w^2)}$



          Now use



          $dfrac{1}{(1-a)}=1+a+a^2....$ formula above






          share|cite|improve this answer














          Hint:



          $dfrac{1}{z^2-8z+25}=dfrac{w^2}{1-(8w-25w^2)}$



          Now use



          $dfrac{1}{(1-a)}=1+a+a^2....$ formula above







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 28 '18 at 12:02

























          answered Nov 28 '18 at 11:48









          Shubham

          1,5951519




          1,5951519












          • But what is the problem with my steps
            – Join_PhD
            Nov 28 '18 at 11:54










          • In this manner, you can approach in less number of steps because you know expansion
            – Shubham
            Nov 28 '18 at 11:55










          • @Shubham: Edit $dfrac{1}{1-a}$.
            – Yadati Kiran
            Nov 28 '18 at 12:00










          • Thanks. Edited @YadatiKiran
            – Shubham
            Nov 28 '18 at 12:03


















          • But what is the problem with my steps
            – Join_PhD
            Nov 28 '18 at 11:54










          • In this manner, you can approach in less number of steps because you know expansion
            – Shubham
            Nov 28 '18 at 11:55










          • @Shubham: Edit $dfrac{1}{1-a}$.
            – Yadati Kiran
            Nov 28 '18 at 12:00










          • Thanks. Edited @YadatiKiran
            – Shubham
            Nov 28 '18 at 12:03
















          But what is the problem with my steps
          – Join_PhD
          Nov 28 '18 at 11:54




          But what is the problem with my steps
          – Join_PhD
          Nov 28 '18 at 11:54












          In this manner, you can approach in less number of steps because you know expansion
          – Shubham
          Nov 28 '18 at 11:55




          In this manner, you can approach in less number of steps because you know expansion
          – Shubham
          Nov 28 '18 at 11:55












          @Shubham: Edit $dfrac{1}{1-a}$.
          – Yadati Kiran
          Nov 28 '18 at 12:00




          @Shubham: Edit $dfrac{1}{1-a}$.
          – Yadati Kiran
          Nov 28 '18 at 12:00












          Thanks. Edited @YadatiKiran
          – Shubham
          Nov 28 '18 at 12:03




          Thanks. Edited @YadatiKiran
          – Shubham
          Nov 28 '18 at 12:03











          1














          Alternatively, notice that
          begin{align}
          frac{1}{z^2-8z+25} &= frac{1}{(z-4+3i)(z-4-3i)} = frac{1}{6i}left(frac{1}{z-4-3i}-frac{1}{z-4+3i}right)\
          &= frac{1}{6i}left(frac{w}{1-w(4+3i)}-frac{w}{1-w(4-3i)}right),
          end{align}
          and then use the geometric series formula on each of the terms.






          share|cite|improve this answer























          • the way what i have done is it correct?
            – Join_PhD
            Nov 28 '18 at 11:55










          • Your steps seem correct. They also don't seem to lead to anything fruitful.
            – MisterRiemann
            Nov 28 '18 at 11:56
















          1














          Alternatively, notice that
          begin{align}
          frac{1}{z^2-8z+25} &= frac{1}{(z-4+3i)(z-4-3i)} = frac{1}{6i}left(frac{1}{z-4-3i}-frac{1}{z-4+3i}right)\
          &= frac{1}{6i}left(frac{w}{1-w(4+3i)}-frac{w}{1-w(4-3i)}right),
          end{align}
          and then use the geometric series formula on each of the terms.






          share|cite|improve this answer























          • the way what i have done is it correct?
            – Join_PhD
            Nov 28 '18 at 11:55










          • Your steps seem correct. They also don't seem to lead to anything fruitful.
            – MisterRiemann
            Nov 28 '18 at 11:56














          1












          1








          1






          Alternatively, notice that
          begin{align}
          frac{1}{z^2-8z+25} &= frac{1}{(z-4+3i)(z-4-3i)} = frac{1}{6i}left(frac{1}{z-4-3i}-frac{1}{z-4+3i}right)\
          &= frac{1}{6i}left(frac{w}{1-w(4+3i)}-frac{w}{1-w(4-3i)}right),
          end{align}
          and then use the geometric series formula on each of the terms.






          share|cite|improve this answer














          Alternatively, notice that
          begin{align}
          frac{1}{z^2-8z+25} &= frac{1}{(z-4+3i)(z-4-3i)} = frac{1}{6i}left(frac{1}{z-4-3i}-frac{1}{z-4+3i}right)\
          &= frac{1}{6i}left(frac{w}{1-w(4+3i)}-frac{w}{1-w(4-3i)}right),
          end{align}
          and then use the geometric series formula on each of the terms.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 28 '18 at 11:55

























          answered Nov 28 '18 at 11:54









          MisterRiemann

          5,7841624




          5,7841624












          • the way what i have done is it correct?
            – Join_PhD
            Nov 28 '18 at 11:55










          • Your steps seem correct. They also don't seem to lead to anything fruitful.
            – MisterRiemann
            Nov 28 '18 at 11:56


















          • the way what i have done is it correct?
            – Join_PhD
            Nov 28 '18 at 11:55










          • Your steps seem correct. They also don't seem to lead to anything fruitful.
            – MisterRiemann
            Nov 28 '18 at 11:56
















          the way what i have done is it correct?
          – Join_PhD
          Nov 28 '18 at 11:55




          the way what i have done is it correct?
          – Join_PhD
          Nov 28 '18 at 11:55












          Your steps seem correct. They also don't seem to lead to anything fruitful.
          – MisterRiemann
          Nov 28 '18 at 11:56




          Your steps seem correct. They also don't seem to lead to anything fruitful.
          – MisterRiemann
          Nov 28 '18 at 11:56











          1














          No, you are not in the right direction. You should perform a partial fractions decomposition:$$frac{w^2}{25left(w^2-frac8{25}w+frac1{25}right)}=frac{frac i6}{frac{4+3i}{25}-w}-frac{frac i6}{frac{4-3i}{25}-w}$$and then apply twice the equality $frac1{1-z}=1+z+z^2+z^2+cdots$






          share|cite|improve this answer





















          • But why is my direction not correct?
            – Join_PhD
            Nov 28 '18 at 11:59










          • I just don't see how you will get the correct answer from that.
            – José Carlos Santos
            Nov 28 '18 at 12:01


















          1














          No, you are not in the right direction. You should perform a partial fractions decomposition:$$frac{w^2}{25left(w^2-frac8{25}w+frac1{25}right)}=frac{frac i6}{frac{4+3i}{25}-w}-frac{frac i6}{frac{4-3i}{25}-w}$$and then apply twice the equality $frac1{1-z}=1+z+z^2+z^2+cdots$






          share|cite|improve this answer





















          • But why is my direction not correct?
            – Join_PhD
            Nov 28 '18 at 11:59










          • I just don't see how you will get the correct answer from that.
            – José Carlos Santos
            Nov 28 '18 at 12:01
















          1












          1








          1






          No, you are not in the right direction. You should perform a partial fractions decomposition:$$frac{w^2}{25left(w^2-frac8{25}w+frac1{25}right)}=frac{frac i6}{frac{4+3i}{25}-w}-frac{frac i6}{frac{4-3i}{25}-w}$$and then apply twice the equality $frac1{1-z}=1+z+z^2+z^2+cdots$






          share|cite|improve this answer












          No, you are not in the right direction. You should perform a partial fractions decomposition:$$frac{w^2}{25left(w^2-frac8{25}w+frac1{25}right)}=frac{frac i6}{frac{4+3i}{25}-w}-frac{frac i6}{frac{4-3i}{25}-w}$$and then apply twice the equality $frac1{1-z}=1+z+z^2+z^2+cdots$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 28 '18 at 11:58









          José Carlos Santos

          151k22123224




          151k22123224












          • But why is my direction not correct?
            – Join_PhD
            Nov 28 '18 at 11:59










          • I just don't see how you will get the correct answer from that.
            – José Carlos Santos
            Nov 28 '18 at 12:01




















          • But why is my direction not correct?
            – Join_PhD
            Nov 28 '18 at 11:59










          • I just don't see how you will get the correct answer from that.
            – José Carlos Santos
            Nov 28 '18 at 12:01


















          But why is my direction not correct?
          – Join_PhD
          Nov 28 '18 at 11:59




          But why is my direction not correct?
          – Join_PhD
          Nov 28 '18 at 11:59












          I just don't see how you will get the correct answer from that.
          – José Carlos Santos
          Nov 28 '18 at 12:01






          I just don't see how you will get the correct answer from that.
          – José Carlos Santos
          Nov 28 '18 at 12:01




















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