Determine the real numbers $a$, $b$, $c$ such that $1$, $frac1{1+omega}$ and $frac1{1+omega^*}$ are zeroes of...












2












$begingroup$


I am stuck on this question:



enter image description here




Let $1$, $omega$ and $omega^*$ be the cube root of unity.



a. Show that $dfrac1{1+omega}=-omega$ and $dfrac1{1+omega^*}=-omega^*$.



b. Determine the real numbers $a$, $b$, $c$ such that $1$, $dfrac1{1+omega}$ and $dfrac1{1+omega^*}$ are zeroes of the polynomial $p(z)=z^3+az^2+bz+c$.



c. Hence, find $p(omega)$ and $p(omega^*)$.




So I was able to do part a by finding the roots in Cartesian forms, but I am not sure how to approach part b.






polynomials complex-numbers roots







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  • $begingroup$
    For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – Martin Sleziak
    7 hours ago
















2












$begingroup$


I am stuck on this question:



enter image description here




Let $1$, $omega$ and $omega^*$ be the cube root of unity.



a. Show that $dfrac1{1+omega}=-omega$ and $dfrac1{1+omega^*}=-omega^*$.



b. Determine the real numbers $a$, $b$, $c$ such that $1$, $dfrac1{1+omega}$ and $dfrac1{1+omega^*}$ are zeroes of the polynomial $p(z)=z^3+az^2+bz+c$.



c. Hence, find $p(omega)$ and $p(omega^*)$.




So I was able to do part a by finding the roots in Cartesian forms, but I am not sure how to approach part b.






polynomials complex-numbers roots







share|cite|improve this question











$endgroup$












  • $begingroup$
    For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – Martin Sleziak
    7 hours ago














2












2








2


1



$begingroup$


I am stuck on this question:



enter image description here




Let $1$, $omega$ and $omega^*$ be the cube root of unity.



a. Show that $dfrac1{1+omega}=-omega$ and $dfrac1{1+omega^*}=-omega^*$.



b. Determine the real numbers $a$, $b$, $c$ such that $1$, $dfrac1{1+omega}$ and $dfrac1{1+omega^*}$ are zeroes of the polynomial $p(z)=z^3+az^2+bz+c$.



c. Hence, find $p(omega)$ and $p(omega^*)$.




So I was able to do part a by finding the roots in Cartesian forms, but I am not sure how to approach part b.






polynomials complex-numbers roots







share|cite|improve this question











$endgroup$




I am stuck on this question:



enter image description here




Let $1$, $omega$ and $omega^*$ be the cube root of unity.



a. Show that $dfrac1{1+omega}=-omega$ and $dfrac1{1+omega^*}=-omega^*$.



b. Determine the real numbers $a$, $b$, $c$ such that $1$, $dfrac1{1+omega}$ and $dfrac1{1+omega^*}$ are zeroes of the polynomial $p(z)=z^3+az^2+bz+c$.



c. Hence, find $p(omega)$ and $p(omega^*)$.




So I was able to do part a by finding the roots in Cartesian forms, but I am not sure how to approach part b.






polynomials complex-numbers roots




polynomials complex-numbers roots






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









Andreas Caranti

57k34397




57k34397










asked 8 hours ago









FluellenFluellen

376




376












  • $begingroup$
    For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – Martin Sleziak
    7 hours ago


















  • $begingroup$
    For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – Martin Sleziak
    7 hours ago
















$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– Martin Sleziak
7 hours ago




$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– Martin Sleziak
7 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

We use the familiar fact that
$$
(x - alpha) (x - beta) = x^{2} - (alpha + beta) x + alpha beta.
$$



For part a, note that $1, omega, omega^{*}$ are the three distinct roots of
$$
x^{3} - 1 = (x -1) (x^{2} + x + 1).
$$
Therefore $omega, omega^{*}$ are the roots of $x^{2} + x + 1$. Hence their product equals the constant coefficient $1$, so that
$$
omega^{*} = omega^{-1},
$$
and their sum equals $-1$, that is, the negative of the coefficient of $x$, so that
$$
omega + omega^{*} = -1,
$$
and your formulas follow.



For b, you may note that because of the above formulas,
$$
frac{1}{1 + omega} = - omega,
quadtext{and}quad
frac{1}{1 + omega^{*}} = - omega^{*}
$$
are the roots of $x^{2} - x + 1$, and so
$$
1, - omega, - omega^{*}
$$
will be the roots of
$$
(x - 1) (x^{2} - x + 1)
=
x^{3} - 2 x^{2} + 2 x - 1.
$$






share|cite|improve this answer









$endgroup$





















    5












    $begingroup$

    For those three numbers to be roots of the cubic equation means that if you set $z$ equal to any one of them, then the cubic is zero. Therefore you can write $$z^3+az^2+bz+cequiv(z-1)left(z-frac1{1+omega}right)left(z-frac1{1+omega^*}right)$$ To determine $a,b,c$, simply multiply this out and equate the coefficients of different powers of $z$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      We use the familiar fact that
      $$
      (x - alpha) (x - beta) = x^{2} - (alpha + beta) x + alpha beta.
      $$



      For part a, note that $1, omega, omega^{*}$ are the three distinct roots of
      $$
      x^{3} - 1 = (x -1) (x^{2} + x + 1).
      $$
      Therefore $omega, omega^{*}$ are the roots of $x^{2} + x + 1$. Hence their product equals the constant coefficient $1$, so that
      $$
      omega^{*} = omega^{-1},
      $$
      and their sum equals $-1$, that is, the negative of the coefficient of $x$, so that
      $$
      omega + omega^{*} = -1,
      $$
      and your formulas follow.



      For b, you may note that because of the above formulas,
      $$
      frac{1}{1 + omega} = - omega,
      quadtext{and}quad
      frac{1}{1 + omega^{*}} = - omega^{*}
      $$
      are the roots of $x^{2} - x + 1$, and so
      $$
      1, - omega, - omega^{*}
      $$
      will be the roots of
      $$
      (x - 1) (x^{2} - x + 1)
      =
      x^{3} - 2 x^{2} + 2 x - 1.
      $$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        We use the familiar fact that
        $$
        (x - alpha) (x - beta) = x^{2} - (alpha + beta) x + alpha beta.
        $$



        For part a, note that $1, omega, omega^{*}$ are the three distinct roots of
        $$
        x^{3} - 1 = (x -1) (x^{2} + x + 1).
        $$
        Therefore $omega, omega^{*}$ are the roots of $x^{2} + x + 1$. Hence their product equals the constant coefficient $1$, so that
        $$
        omega^{*} = omega^{-1},
        $$
        and their sum equals $-1$, that is, the negative of the coefficient of $x$, so that
        $$
        omega + omega^{*} = -1,
        $$
        and your formulas follow.



        For b, you may note that because of the above formulas,
        $$
        frac{1}{1 + omega} = - omega,
        quadtext{and}quad
        frac{1}{1 + omega^{*}} = - omega^{*}
        $$
        are the roots of $x^{2} - x + 1$, and so
        $$
        1, - omega, - omega^{*}
        $$
        will be the roots of
        $$
        (x - 1) (x^{2} - x + 1)
        =
        x^{3} - 2 x^{2} + 2 x - 1.
        $$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          We use the familiar fact that
          $$
          (x - alpha) (x - beta) = x^{2} - (alpha + beta) x + alpha beta.
          $$



          For part a, note that $1, omega, omega^{*}$ are the three distinct roots of
          $$
          x^{3} - 1 = (x -1) (x^{2} + x + 1).
          $$
          Therefore $omega, omega^{*}$ are the roots of $x^{2} + x + 1$. Hence their product equals the constant coefficient $1$, so that
          $$
          omega^{*} = omega^{-1},
          $$
          and their sum equals $-1$, that is, the negative of the coefficient of $x$, so that
          $$
          omega + omega^{*} = -1,
          $$
          and your formulas follow.



          For b, you may note that because of the above formulas,
          $$
          frac{1}{1 + omega} = - omega,
          quadtext{and}quad
          frac{1}{1 + omega^{*}} = - omega^{*}
          $$
          are the roots of $x^{2} - x + 1$, and so
          $$
          1, - omega, - omega^{*}
          $$
          will be the roots of
          $$
          (x - 1) (x^{2} - x + 1)
          =
          x^{3} - 2 x^{2} + 2 x - 1.
          $$






          share|cite|improve this answer









          $endgroup$



          We use the familiar fact that
          $$
          (x - alpha) (x - beta) = x^{2} - (alpha + beta) x + alpha beta.
          $$



          For part a, note that $1, omega, omega^{*}$ are the three distinct roots of
          $$
          x^{3} - 1 = (x -1) (x^{2} + x + 1).
          $$
          Therefore $omega, omega^{*}$ are the roots of $x^{2} + x + 1$. Hence their product equals the constant coefficient $1$, so that
          $$
          omega^{*} = omega^{-1},
          $$
          and their sum equals $-1$, that is, the negative of the coefficient of $x$, so that
          $$
          omega + omega^{*} = -1,
          $$
          and your formulas follow.



          For b, you may note that because of the above formulas,
          $$
          frac{1}{1 + omega} = - omega,
          quadtext{and}quad
          frac{1}{1 + omega^{*}} = - omega^{*}
          $$
          are the roots of $x^{2} - x + 1$, and so
          $$
          1, - omega, - omega^{*}
          $$
          will be the roots of
          $$
          (x - 1) (x^{2} - x + 1)
          =
          x^{3} - 2 x^{2} + 2 x - 1.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 7 hours ago









          Andreas CarantiAndreas Caranti

          57k34397




          57k34397























              5












              $begingroup$

              For those three numbers to be roots of the cubic equation means that if you set $z$ equal to any one of them, then the cubic is zero. Therefore you can write $$z^3+az^2+bz+cequiv(z-1)left(z-frac1{1+omega}right)left(z-frac1{1+omega^*}right)$$ To determine $a,b,c$, simply multiply this out and equate the coefficients of different powers of $z$.






              share|cite|improve this answer









              $endgroup$


















                5












                $begingroup$

                For those three numbers to be roots of the cubic equation means that if you set $z$ equal to any one of them, then the cubic is zero. Therefore you can write $$z^3+az^2+bz+cequiv(z-1)left(z-frac1{1+omega}right)left(z-frac1{1+omega^*}right)$$ To determine $a,b,c$, simply multiply this out and equate the coefficients of different powers of $z$.






                share|cite|improve this answer









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  For those three numbers to be roots of the cubic equation means that if you set $z$ equal to any one of them, then the cubic is zero. Therefore you can write $$z^3+az^2+bz+cequiv(z-1)left(z-frac1{1+omega}right)left(z-frac1{1+omega^*}right)$$ To determine $a,b,c$, simply multiply this out and equate the coefficients of different powers of $z$.






                  share|cite|improve this answer









                  $endgroup$



                  For those three numbers to be roots of the cubic equation means that if you set $z$ equal to any one of them, then the cubic is zero. Therefore you can write $$z^3+az^2+bz+cequiv(z-1)left(z-frac1{1+omega}right)left(z-frac1{1+omega^*}right)$$ To determine $a,b,c$, simply multiply this out and equate the coefficients of different powers of $z$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 8 hours ago









                  John DoeJohn Doe

                  11.8k11239




                  11.8k11239






























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