Show that z is purely imaginary

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Question



Show that z is a purely imaginary number given that



$(z+1)^100 = (z-1)^100$



Rearrange to



$(fracz+1z-1)^100 = 1$



I tried using $z = x+iy$ and trying to multiply numerator and denominator by the conjugate, but I hit a roadblock. I also tried substituting $1 = -e^ipi$, but that also doesn't seem to get me anywhere. How can I prove this? Any help is appreciated, thank you!






complex-analysis complex-numbers proof-writing







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    Question



    Show that z is a purely imaginary number given that



    $(z+1)^100 = (z-1)^100$



    Rearrange to



    $(fracz+1z-1)^100 = 1$



    I tried using $z = x+iy$ and trying to multiply numerator and denominator by the conjugate, but I hit a roadblock. I also tried substituting $1 = -e^ipi$, but that also doesn't seem to get me anywhere. How can I prove this? Any help is appreciated, thank you!






    complex-analysis complex-numbers proof-writing







    share|cite|improve this question









    New contributor




    Chinmayee Gidwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





















      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Question



      Show that z is a purely imaginary number given that



      $(z+1)^100 = (z-1)^100$



      Rearrange to



      $(fracz+1z-1)^100 = 1$



      I tried using $z = x+iy$ and trying to multiply numerator and denominator by the conjugate, but I hit a roadblock. I also tried substituting $1 = -e^ipi$, but that also doesn't seem to get me anywhere. How can I prove this? Any help is appreciated, thank you!






      complex-analysis complex-numbers proof-writing







      share|cite|improve this question









      New contributor




      Chinmayee Gidwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      Question



      Show that z is a purely imaginary number given that



      $(z+1)^100 = (z-1)^100$



      Rearrange to



      $(fracz+1z-1)^100 = 1$



      I tried using $z = x+iy$ and trying to multiply numerator and denominator by the conjugate, but I hit a roadblock. I also tried substituting $1 = -e^ipi$, but that also doesn't seem to get me anywhere. How can I prove this? Any help is appreciated, thank you!






      complex-analysis complex-numbers proof-writing




      complex-analysis complex-numbers proof-writing






      share|cite|improve this question









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      Chinmayee Gidwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









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      share|cite|improve this question








      edited 2 hours ago









      abc...

      2,134529




      2,134529






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      asked 2 hours ago









      Chinmayee Gidwani

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          3 Answers
          3






          active

          oldest

          votes

















          up vote
          4
          down vote



          accepted










          Hint
          $$|z+1|^2=|z-1|^2 \
          left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
          z barz+z+barz+1=zbarz-z-barz+1 \
          barz=-z
          $$






          share|cite|improve this answer




















          • Umm does $overline z$ stand for the complex conjungate of $z$?
            – Mohammad Zuhair Khan
            2 hours ago










          • @MohammadZuhairKhan Yes.
            – N. S.
            2 hours ago

















          up vote
          1
          down vote













          Take absolute values, $|z+1|^100=|z-1|^100|$. Absolute values are always no negative real numbers, for which root extractions are always defined and one-to-one functions; so take $100^th$ roots to get $|z+1|=|z-1|$. Oops, I have an urgent call and need you to continue from there ;-) .






          share|cite|improve this answer




















          • is it valid to take the abs value of both sides like that?
            – Chinmayee Gidwani
            1 hour ago










          • Yup, if two elements are equal so are their absolute vaues.
            – Oscar Lanzi
            1 hour ago

















          up vote
          1
          down vote














          Note that $|z+1|=|z-1|$




          Therefore



          $sqrt (x+1)^2+y^2=sqrt (x-1)^2+y^2$



          $(x+1)^2+y^2=(x-1)^2+y^2$



          $x^2+2x+1=x^2-2x+1$



          $4x=0$



          $therefore x=0$






          share|cite|improve this answer




















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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            4
            down vote



            accepted










            Hint
            $$|z+1|^2=|z-1|^2 \
            left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
            z barz+z+barz+1=zbarz-z-barz+1 \
            barz=-z
            $$






            share|cite|improve this answer




















            • Umm does $overline z$ stand for the complex conjungate of $z$?
              – Mohammad Zuhair Khan
              2 hours ago










            • @MohammadZuhairKhan Yes.
              – N. S.
              2 hours ago














            up vote
            4
            down vote



            accepted










            Hint
            $$|z+1|^2=|z-1|^2 \
            left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
            z barz+z+barz+1=zbarz-z-barz+1 \
            barz=-z
            $$






            share|cite|improve this answer




















            • Umm does $overline z$ stand for the complex conjungate of $z$?
              – Mohammad Zuhair Khan
              2 hours ago










            • @MohammadZuhairKhan Yes.
              – N. S.
              2 hours ago












            up vote
            4
            down vote



            accepted







            up vote
            4
            down vote



            accepted






            Hint
            $$|z+1|^2=|z-1|^2 \
            left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
            z barz+z+barz+1=zbarz-z-barz+1 \
            barz=-z
            $$






            share|cite|improve this answer












            Hint
            $$|z+1|^2=|z-1|^2 \
            left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
            z barz+z+barz+1=zbarz-z-barz+1 \
            barz=-z
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 hours ago









            N. S.

            98.8k5106197




            98.8k5106197











            • Umm does $overline z$ stand for the complex conjungate of $z$?
              – Mohammad Zuhair Khan
              2 hours ago










            • @MohammadZuhairKhan Yes.
              – N. S.
              2 hours ago
















            • Umm does $overline z$ stand for the complex conjungate of $z$?
              – Mohammad Zuhair Khan
              2 hours ago










            • @MohammadZuhairKhan Yes.
              – N. S.
              2 hours ago















            Umm does $overline z$ stand for the complex conjungate of $z$?
            – Mohammad Zuhair Khan
            2 hours ago




            Umm does $overline z$ stand for the complex conjungate of $z$?
            – Mohammad Zuhair Khan
            2 hours ago












            @MohammadZuhairKhan Yes.
            – N. S.
            2 hours ago




            @MohammadZuhairKhan Yes.
            – N. S.
            2 hours ago










            up vote
            1
            down vote













            Take absolute values, $|z+1|^100=|z-1|^100|$. Absolute values are always no negative real numbers, for which root extractions are always defined and one-to-one functions; so take $100^th$ roots to get $|z+1|=|z-1|$. Oops, I have an urgent call and need you to continue from there ;-) .






            share|cite|improve this answer




















            • is it valid to take the abs value of both sides like that?
              – Chinmayee Gidwani
              1 hour ago










            • Yup, if two elements are equal so are their absolute vaues.
              – Oscar Lanzi
              1 hour ago














            up vote
            1
            down vote













            Take absolute values, $|z+1|^100=|z-1|^100|$. Absolute values are always no negative real numbers, for which root extractions are always defined and one-to-one functions; so take $100^th$ roots to get $|z+1|=|z-1|$. Oops, I have an urgent call and need you to continue from there ;-) .






            share|cite|improve this answer




















            • is it valid to take the abs value of both sides like that?
              – Chinmayee Gidwani
              1 hour ago










            • Yup, if two elements are equal so are their absolute vaues.
              – Oscar Lanzi
              1 hour ago












            up vote
            1
            down vote










            up vote
            1
            down vote









            Take absolute values, $|z+1|^100=|z-1|^100|$. Absolute values are always no negative real numbers, for which root extractions are always defined and one-to-one functions; so take $100^th$ roots to get $|z+1|=|z-1|$. Oops, I have an urgent call and need you to continue from there ;-) .






            share|cite|improve this answer












            Take absolute values, $|z+1|^100=|z-1|^100|$. Absolute values are always no negative real numbers, for which root extractions are always defined and one-to-one functions; so take $100^th$ roots to get $|z+1|=|z-1|$. Oops, I have an urgent call and need you to continue from there ;-) .







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 hours ago









            Oscar Lanzi

            10.4k11733




            10.4k11733











            • is it valid to take the abs value of both sides like that?
              – Chinmayee Gidwani
              1 hour ago










            • Yup, if two elements are equal so are their absolute vaues.
              – Oscar Lanzi
              1 hour ago
















            • is it valid to take the abs value of both sides like that?
              – Chinmayee Gidwani
              1 hour ago










            • Yup, if two elements are equal so are their absolute vaues.
              – Oscar Lanzi
              1 hour ago















            is it valid to take the abs value of both sides like that?
            – Chinmayee Gidwani
            1 hour ago




            is it valid to take the abs value of both sides like that?
            – Chinmayee Gidwani
            1 hour ago












            Yup, if two elements are equal so are their absolute vaues.
            – Oscar Lanzi
            1 hour ago




            Yup, if two elements are equal so are their absolute vaues.
            – Oscar Lanzi
            1 hour ago










            up vote
            1
            down vote














            Note that $|z+1|=|z-1|$




            Therefore



            $sqrt (x+1)^2+y^2=sqrt (x-1)^2+y^2$



            $(x+1)^2+y^2=(x-1)^2+y^2$



            $x^2+2x+1=x^2-2x+1$



            $4x=0$



            $therefore x=0$






            share|cite|improve this answer
























              up vote
              1
              down vote














              Note that $|z+1|=|z-1|$




              Therefore



              $sqrt (x+1)^2+y^2=sqrt (x-1)^2+y^2$



              $(x+1)^2+y^2=(x-1)^2+y^2$



              $x^2+2x+1=x^2-2x+1$



              $4x=0$



              $therefore x=0$






              share|cite|improve this answer






















                up vote
                1
                down vote










                up vote
                1
                down vote










                Note that $|z+1|=|z-1|$




                Therefore



                $sqrt (x+1)^2+y^2=sqrt (x-1)^2+y^2$



                $(x+1)^2+y^2=(x-1)^2+y^2$



                $x^2+2x+1=x^2-2x+1$



                $4x=0$



                $therefore x=0$






                share|cite|improve this answer













                Note that $|z+1|=|z-1|$




                Therefore



                $sqrt (x+1)^2+y^2=sqrt (x-1)^2+y^2$



                $(x+1)^2+y^2=(x-1)^2+y^2$



                $x^2+2x+1=x^2-2x+1$



                $4x=0$



                $therefore x=0$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                Mohammad Zuhair Khan

                918422




                918422




















                    Chinmayee Gidwani is a new contributor. Be nice, and check out our Code of Conduct.









                     

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