Show that if $(z+1)^100 = (z-1)^100$, then $z$ is purely imaginary

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Let $z$ be a complex number satisfying
$$(z+1)^100 = (z-1)^100$$
Show that $z$ is purely imaginary, i.e. that $Re(z) = 0$.




Rearrange to



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



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







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    up vote
    7
    down vote

    favorite













    Let $z$ be a complex number satisfying
    $$(z+1)^100 = (z-1)^100$$
    Show that $z$ is purely imaginary, i.e. that $Re(z) = 0$.




    Rearrange to



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



    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







    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
      7
      down vote

      favorite









      up vote
      7
      down vote

      favorite












      Let $z$ be a complex number satisfying
      $$(z+1)^100 = (z-1)^100$$
      Show that $z$ is purely imaginary, i.e. that $Re(z) = 0$.




      Rearrange to



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



      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







      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.












      Let $z$ be a complex number satisfying
      $$(z+1)^100 = (z-1)^100$$
      Show that $z$ is purely imaginary, i.e. that $Re(z) = 0$.




      Rearrange to



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



      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




      complex-analysis complex-numbers






      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.











      share|cite|improve this question









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      edited 3 mins ago









      TheSimpliFire

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      asked 7 hours ago









      Chinmayee Gidwani

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          3 Answers
          3






          active

          oldest

          votes

















          up vote
          10
          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
            7 hours ago










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

















          up vote
          4
          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
            6 hours ago










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

















          up vote
          3
          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




















          • So it is true for all $z in i mathbb R$?
            – CompuChip
            39 mins ago










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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          10
          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
            7 hours ago










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














          up vote
          10
          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
            7 hours ago










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












          up vote
          10
          down vote



          accepted







          up vote
          10
          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 7 hours ago









          N. S.

          98.8k5106197




          98.8k5106197











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










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
















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










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















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




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












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




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










          up vote
          4
          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
            6 hours ago










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














          up vote
          4
          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
            6 hours ago










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












          up vote
          4
          down vote










          up vote
          4
          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 7 hours ago









          Oscar Lanzi

          10.4k11733




          10.4k11733











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










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
















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










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















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




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












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




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










          up vote
          3
          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




















          • So it is true for all $z in i mathbb R$?
            – CompuChip
            39 mins ago














          up vote
          3
          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




















          • So it is true for all $z in i mathbb R$?
            – CompuChip
            39 mins ago












          up vote
          3
          down vote










          up vote
          3
          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 7 hours ago









          Mohammad Zuhair Khan

          938422




          938422











          • So it is true for all $z in i mathbb R$?
            – CompuChip
            39 mins ago
















          • So it is true for all $z in i mathbb R$?
            – CompuChip
            39 mins ago















          So it is true for all $z in i mathbb R$?
          – CompuChip
          39 mins ago




          So it is true for all $z in i mathbb R$?
          – CompuChip
          39 mins ago










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









           

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