Poles of $(e^z+1)^-1$

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So I know the function $(e^z+1)^-1$ has a pole at the points $(1+2n)pi i$. How do I show this pole is simple (i.e order 1)? I have attempted to multiply the above by $(z- i pi)$ and show that the resulting function is holomorphic, however that has not been particularly fruitful. Does anyone have any advice? Will I have to compute a Laurent series? If so, how would I do that?




complex-analysis




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  • L'Hôpital's rule works.
    – saulspatz
    Jul 16 at 3:53














up vote
1
down vote

favorite












So I know the function $(e^z+1)^-1$ has a pole at the points $(1+2n)pi i$. How do I show this pole is simple (i.e order 1)? I have attempted to multiply the above by $(z- i pi)$ and show that the resulting function is holomorphic, however that has not been particularly fruitful. Does anyone have any advice? Will I have to compute a Laurent series? If so, how would I do that?




complex-analysis




share|cite|improve this question





















  • L'Hôpital's rule works.
    – saulspatz
    Jul 16 at 3:53












up vote
1
down vote

favorite









up vote
1
down vote

favorite











So I know the function $(e^z+1)^-1$ has a pole at the points $(1+2n)pi i$. How do I show this pole is simple (i.e order 1)? I have attempted to multiply the above by $(z- i pi)$ and show that the resulting function is holomorphic, however that has not been particularly fruitful. Does anyone have any advice? Will I have to compute a Laurent series? If so, how would I do that?




complex-analysis




share|cite|improve this question













So I know the function $(e^z+1)^-1$ has a pole at the points $(1+2n)pi i$. How do I show this pole is simple (i.e order 1)? I have attempted to multiply the above by $(z- i pi)$ and show that the resulting function is holomorphic, however that has not been particularly fruitful. Does anyone have any advice? Will I have to compute a Laurent series? If so, how would I do that?





complex-analysis





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edited Jul 16 at 4:30









Parcly Taxel

33.6k136588




33.6k136588









asked Jul 16 at 3:46









rubikscube09

869617




869617











  • L'Hôpital's rule works.
    – saulspatz
    Jul 16 at 3:53
















  • L'Hôpital's rule works.
    – saulspatz
    Jul 16 at 3:53















L'Hôpital's rule works.
– saulspatz
Jul 16 at 3:53




L'Hôpital's rule works.
– saulspatz
Jul 16 at 3:53










4 Answers
4






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up vote
4
down vote



accepted










Theorem. If $p$ and $q$ are holomorphic at $z_0$ and $p(z_0)ne0$ and $q(z_0)=0$ and $q'(z_0)ne0$, then $p/q$ has a simple pole at $z_0$.






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  • I assume this can be proven using the argument principle?
    – rubikscube09
    Jul 16 at 4:00










  • I would just use L'Hopital.
    – David
    Jul 16 at 4:13










  • @rubikscube09 alternatively, use that you can write $q(z) = (z-z_0)f$, where $f$ is holomorphic and nonvanishing at $z_0$, then that $1/f$ is again holomorphic.
    – Mike Miller
    Jul 16 at 5:04

















up vote
1
down vote













Why not fruitful ?? Let $f(z)=(e^z+1)^-1$ , $g(z)=e^z+1$ and $z_n=(1+2n)pi i$ for $n in mathbb Z$.



Then $(z-z_n)f(z)= fracz-z_ng(z)-g(z_n) to g'(z_n)^-1=-1 ne 0$ as $z to z_n$.



This shows that each $z_n$ is a simple pole of $f$.






share|cite|improve this answer




























    up vote
    1
    down vote













    You have that $fracz-ipie^z+1 $ is the reciprocal of a difference quotient. Take the limit of this difference quotient to see that you get the derivative of $e^z$ at $ipi$ which is simply $e^ipi$ since $e^z$ is its own derivative.






    share|cite|improve this answer




























      up vote
      0
      down vote













      The function has a pole iff it is a zero of $e^z+1$.



      Note that the derivative is $e^z$ hence any zero is simple hence the pole is simple.






      share|cite|improve this answer





















      • Why does the derivative being $e^z$ guarantee the pole is simple?
        – rubikscube09
        Jul 16 at 4:25










      • If the pole had order greater than one then the zero would have order greater than one and hence the derivative would have a zero there too.
        – copper.hat
        Jul 16 at 4:32










      Your Answer




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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      4
      down vote



      accepted










      Theorem. If $p$ and $q$ are holomorphic at $z_0$ and $p(z_0)ne0$ and $q(z_0)=0$ and $q'(z_0)ne0$, then $p/q$ has a simple pole at $z_0$.






      share|cite|improve this answer





















      • I assume this can be proven using the argument principle?
        – rubikscube09
        Jul 16 at 4:00










      • I would just use L'Hopital.
        – David
        Jul 16 at 4:13










      • @rubikscube09 alternatively, use that you can write $q(z) = (z-z_0)f$, where $f$ is holomorphic and nonvanishing at $z_0$, then that $1/f$ is again holomorphic.
        – Mike Miller
        Jul 16 at 5:04














      up vote
      4
      down vote



      accepted










      Theorem. If $p$ and $q$ are holomorphic at $z_0$ and $p(z_0)ne0$ and $q(z_0)=0$ and $q'(z_0)ne0$, then $p/q$ has a simple pole at $z_0$.






      share|cite|improve this answer





















      • I assume this can be proven using the argument principle?
        – rubikscube09
        Jul 16 at 4:00










      • I would just use L'Hopital.
        – David
        Jul 16 at 4:13










      • @rubikscube09 alternatively, use that you can write $q(z) = (z-z_0)f$, where $f$ is holomorphic and nonvanishing at $z_0$, then that $1/f$ is again holomorphic.
        – Mike Miller
        Jul 16 at 5:04












      up vote
      4
      down vote



      accepted







      up vote
      4
      down vote



      accepted






      Theorem. If $p$ and $q$ are holomorphic at $z_0$ and $p(z_0)ne0$ and $q(z_0)=0$ and $q'(z_0)ne0$, then $p/q$ has a simple pole at $z_0$.






      share|cite|improve this answer













      Theorem. If $p$ and $q$ are holomorphic at $z_0$ and $p(z_0)ne0$ and $q(z_0)=0$ and $q'(z_0)ne0$, then $p/q$ has a simple pole at $z_0$.







      share|cite|improve this answer













      share|cite|improve this answer



      share|cite|improve this answer











      answered Jul 16 at 3:54









      David

      66k662125




      66k662125











      • I assume this can be proven using the argument principle?
        – rubikscube09
        Jul 16 at 4:00










      • I would just use L'Hopital.
        – David
        Jul 16 at 4:13










      • @rubikscube09 alternatively, use that you can write $q(z) = (z-z_0)f$, where $f$ is holomorphic and nonvanishing at $z_0$, then that $1/f$ is again holomorphic.
        – Mike Miller
        Jul 16 at 5:04
















      • I assume this can be proven using the argument principle?
        – rubikscube09
        Jul 16 at 4:00










      • I would just use L'Hopital.
        – David
        Jul 16 at 4:13










      • @rubikscube09 alternatively, use that you can write $q(z) = (z-z_0)f$, where $f$ is holomorphic and nonvanishing at $z_0$, then that $1/f$ is again holomorphic.
        – Mike Miller
        Jul 16 at 5:04















      I assume this can be proven using the argument principle?
      – rubikscube09
      Jul 16 at 4:00




      I assume this can be proven using the argument principle?
      – rubikscube09
      Jul 16 at 4:00












      I would just use L'Hopital.
      – David
      Jul 16 at 4:13




      I would just use L'Hopital.
      – David
      Jul 16 at 4:13












      @rubikscube09 alternatively, use that you can write $q(z) = (z-z_0)f$, where $f$ is holomorphic and nonvanishing at $z_0$, then that $1/f$ is again holomorphic.
      – Mike Miller
      Jul 16 at 5:04




      @rubikscube09 alternatively, use that you can write $q(z) = (z-z_0)f$, where $f$ is holomorphic and nonvanishing at $z_0$, then that $1/f$ is again holomorphic.
      – Mike Miller
      Jul 16 at 5:04










      up vote
      1
      down vote













      Why not fruitful ?? Let $f(z)=(e^z+1)^-1$ , $g(z)=e^z+1$ and $z_n=(1+2n)pi i$ for $n in mathbb Z$.



      Then $(z-z_n)f(z)= fracz-z_ng(z)-g(z_n) to g'(z_n)^-1=-1 ne 0$ as $z to z_n$.



      This shows that each $z_n$ is a simple pole of $f$.






      share|cite|improve this answer

























        up vote
        1
        down vote













        Why not fruitful ?? Let $f(z)=(e^z+1)^-1$ , $g(z)=e^z+1$ and $z_n=(1+2n)pi i$ for $n in mathbb Z$.



        Then $(z-z_n)f(z)= fracz-z_ng(z)-g(z_n) to g'(z_n)^-1=-1 ne 0$ as $z to z_n$.



        This shows that each $z_n$ is a simple pole of $f$.






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          Why not fruitful ?? Let $f(z)=(e^z+1)^-1$ , $g(z)=e^z+1$ and $z_n=(1+2n)pi i$ for $n in mathbb Z$.



          Then $(z-z_n)f(z)= fracz-z_ng(z)-g(z_n) to g'(z_n)^-1=-1 ne 0$ as $z to z_n$.



          This shows that each $z_n$ is a simple pole of $f$.






          share|cite|improve this answer













          Why not fruitful ?? Let $f(z)=(e^z+1)^-1$ , $g(z)=e^z+1$ and $z_n=(1+2n)pi i$ for $n in mathbb Z$.



          Then $(z-z_n)f(z)= fracz-z_ng(z)-g(z_n) to g'(z_n)^-1=-1 ne 0$ as $z to z_n$.



          This shows that each $z_n$ is a simple pole of $f$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 16 at 4:40









          Fred

          37.6k1237




          37.6k1237




















              up vote
              1
              down vote













              You have that $fracz-ipie^z+1 $ is the reciprocal of a difference quotient. Take the limit of this difference quotient to see that you get the derivative of $e^z$ at $ipi$ which is simply $e^ipi$ since $e^z$ is its own derivative.






              share|cite|improve this answer

























                up vote
                1
                down vote













                You have that $fracz-ipie^z+1 $ is the reciprocal of a difference quotient. Take the limit of this difference quotient to see that you get the derivative of $e^z$ at $ipi$ which is simply $e^ipi$ since $e^z$ is its own derivative.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  You have that $fracz-ipie^z+1 $ is the reciprocal of a difference quotient. Take the limit of this difference quotient to see that you get the derivative of $e^z$ at $ipi$ which is simply $e^ipi$ since $e^z$ is its own derivative.






                  share|cite|improve this answer













                  You have that $fracz-ipie^z+1 $ is the reciprocal of a difference quotient. Take the limit of this difference quotient to see that you get the derivative of $e^z$ at $ipi$ which is simply $e^ipi$ since $e^z$ is its own derivative.







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Jul 17 at 2:01









                  Josh

                  477




                  477




















                      up vote
                      0
                      down vote













                      The function has a pole iff it is a zero of $e^z+1$.



                      Note that the derivative is $e^z$ hence any zero is simple hence the pole is simple.






                      share|cite|improve this answer





















                      • Why does the derivative being $e^z$ guarantee the pole is simple?
                        – rubikscube09
                        Jul 16 at 4:25










                      • If the pole had order greater than one then the zero would have order greater than one and hence the derivative would have a zero there too.
                        – copper.hat
                        Jul 16 at 4:32














                      up vote
                      0
                      down vote













                      The function has a pole iff it is a zero of $e^z+1$.



                      Note that the derivative is $e^z$ hence any zero is simple hence the pole is simple.






                      share|cite|improve this answer





















                      • Why does the derivative being $e^z$ guarantee the pole is simple?
                        – rubikscube09
                        Jul 16 at 4:25










                      • If the pole had order greater than one then the zero would have order greater than one and hence the derivative would have a zero there too.
                        – copper.hat
                        Jul 16 at 4:32












                      up vote
                      0
                      down vote










                      up vote
                      0
                      down vote









                      The function has a pole iff it is a zero of $e^z+1$.



                      Note that the derivative is $e^z$ hence any zero is simple hence the pole is simple.






                      share|cite|improve this answer













                      The function has a pole iff it is a zero of $e^z+1$.



                      Note that the derivative is $e^z$ hence any zero is simple hence the pole is simple.







                      share|cite|improve this answer













                      share|cite|improve this answer



                      share|cite|improve this answer











                      answered Jul 16 at 4:19









                      copper.hat

                      122k557156




                      122k557156











                      • Why does the derivative being $e^z$ guarantee the pole is simple?
                        – rubikscube09
                        Jul 16 at 4:25










                      • If the pole had order greater than one then the zero would have order greater than one and hence the derivative would have a zero there too.
                        – copper.hat
                        Jul 16 at 4:32
















                      • Why does the derivative being $e^z$ guarantee the pole is simple?
                        – rubikscube09
                        Jul 16 at 4:25










                      • If the pole had order greater than one then the zero would have order greater than one and hence the derivative would have a zero there too.
                        – copper.hat
                        Jul 16 at 4:32















                      Why does the derivative being $e^z$ guarantee the pole is simple?
                      – rubikscube09
                      Jul 16 at 4:25




                      Why does the derivative being $e^z$ guarantee the pole is simple?
                      – rubikscube09
                      Jul 16 at 4:25












                      If the pole had order greater than one then the zero would have order greater than one and hence the derivative would have a zero there too.
                      – copper.hat
                      Jul 16 at 4:32




                      If the pole had order greater than one then the zero would have order greater than one and hence the derivative would have a zero there too.
                      – copper.hat
                      Jul 16 at 4:32












                       

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