Determine the real numbers $a$, $b$, $c$ such that $1$, $frac11+omega$ and $frac11+omega^*$ are zeroes of the polynomial $p(z)=z^3+az^2+bz+c$Determine the number of real roots in the equation…Finding a polynomial with product and sum of its zeroesGiven roots (real and complex), find the polynomialWhat are the conditions on $a$ such that the polynomial $x^4-2ax^2+x+a^2-a$ has four distinct real roots?Zeroes of the polynomial $f(x)$ over the field $F$ of order 256.Perturbing coefficients of a polynomial with simple real zeroesApart from the Fundamental Theorem of Algebra and Descartes Rule of Signs, are there any other ways to determine the nature of roots of a polynomial?How the determine the number of real positive roots of a polynomial?Finding the real roots and complex roots to a polynomialZeroes of a sextic polynomial and analytical form
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Determine the real numbers $a$, $b$, $c$ such that $1$, $frac11+omega$ and $frac11+omega^*$ are zeroes of the polynomial $p(z)=z^3+az^2+bz+c$
Determine the number of real roots in the equation…Finding a polynomial with product and sum of its zeroesGiven roots (real and complex), find the polynomialWhat are the conditions on $a$ such that the polynomial $x^4-2ax^2+x+a^2-a$ has four distinct real roots?Zeroes of the polynomial $f(x)$ over the field $F$ of order 256.Perturbing coefficients of a polynomial with simple real zeroesApart from the Fundamental Theorem of Algebra and Descartes Rule of Signs, are there any other ways to determine the nature of roots of a polynomial?How the determine the number of real positive roots of a polynomial?Finding the real roots and complex roots to a polynomialZeroes of a sextic polynomial and analytical form
$begingroup$
I am stuck on this question:
Let $1$, $omega$ and $omega^*$ be the cube root of unity.
a. Show that $dfrac11+omega=-omega$ and $dfrac11+omega^*=-omega^*$.
b. Determine the real numbers $a$, $b$, $c$ such that $1$, $dfrac11+omega$ and $dfrac11+omega^*$ 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
edited 6 hours ago
Andreas Caranti
57k34397
asked 10 hours ago
FluellenFluellen
376
$endgroup$
add a comment |
$begingroup$
I am stuck on this question:
Let $1$, $omega$ and $omega^*$ be the cube root of unity.
a. Show that $dfrac11+omega=-omega$ and $dfrac11+omega^*=-omega^*$.
b. Determine the real numbers $a$, $b$, $c$ such that $1$, $dfrac11+omega$ and $dfrac11+omega^*$ 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
edited 6 hours ago
Andreas Caranti
57k34397
asked 10 hours ago
FluellenFluellen
376
$endgroup$
$begingroup$
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
9 hours ago
add a comment |
$begingroup$
I am stuck on this question:
Let $1$, $omega$ and $omega^*$ be the cube root of unity.
a. Show that $dfrac11+omega=-omega$ and $dfrac11+omega^*=-omega^*$.
b. Determine the real numbers $a$, $b$, $c$ such that $1$, $dfrac11+omega$ and $dfrac11+omega^*$ 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
edited 6 hours ago
Andreas Caranti
57k34397
asked 10 hours ago
FluellenFluellen
376
$endgroup$
I am stuck on this question:
Let $1$, $omega$ and $omega^*$ be the cube root of unity.
a. Show that $dfrac11+omega=-omega$ and $dfrac11+omega^*=-omega^*$.
b. Determine the real numbers $a$, $b$, $c$ such that $1$, $dfrac11+omega$ and $dfrac11+omega^*$ 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
edited 6 hours ago
Andreas Caranti
57k34397
asked 10 hours ago
FluellenFluellen
376
edited 6 hours ago
Andreas Caranti
57k34397
asked 10 hours ago
FluellenFluellen
376
edited 6 hours ago
Andreas Caranti
57k34397
edited 6 hours ago
Andreas Caranti
57k34397
edited 6 hours ago
Andreas Caranti
57k34397
57k34397
asked 10 hours ago
FluellenFluellen
376
asked 10 hours ago
FluellenFluellen
376
asked 10 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
9 hours ago
add a comment |
$begingroup$
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
9 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
9 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
9 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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,
$$
frac11 + omega = - omega,
quadtextandquad
frac11 + omega^* = - omega^*
$$
are the roots of $x^2 - x + 1$, and so
$$
1, - omega, - omega^*
$$
will be the roots of
$$
(x - 1) (x^2 - x + 1)
=
x^3 - 2 x^2 + 2 x - 1.
$$
answered 10 hours ago
Andreas CarantiAndreas Caranti
57k34397
$endgroup$
add a comment |
$begingroup$
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 $$z^3+az^2+bz+cequiv(z-1)left(z-frac11+omegaright)left(z-frac11+omega^*right)$$ To determine $a,b,c$, simply multiply this out and equate the coefficients of different powers of $z$.
answered 10 hours ago
John DoeJohn Doe
11.8k11239
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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,
$$
frac11 + omega = - omega,
quadtextandquad
frac11 + omega^* = - omega^*
$$
are the roots of $x^2 - x + 1$, and so
$$
1, - omega, - omega^*
$$
will be the roots of
$$
(x - 1) (x^2 - x + 1)
=
x^3 - 2 x^2 + 2 x - 1.
$$
answered 10 hours ago
Andreas CarantiAndreas Caranti
57k34397
$endgroup$
add a comment |
$begingroup$
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,
$$
frac11 + omega = - omega,
quadtextandquad
frac11 + omega^* = - omega^*
$$
are the roots of $x^2 - x + 1$, and so
$$
1, - omega, - omega^*
$$
will be the roots of
$$
(x - 1) (x^2 - x + 1)
=
x^3 - 2 x^2 + 2 x - 1.
$$
answered 10 hours ago
Andreas CarantiAndreas Caranti
57k34397
$endgroup$
add a comment |
$begingroup$
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,
$$
frac11 + omega = - omega,
quadtextandquad
frac11 + omega^* = - omega^*
$$
are the roots of $x^2 - x + 1$, and so
$$
1, - omega, - omega^*
$$
will be the roots of
$$
(x - 1) (x^2 - x + 1)
=
x^3 - 2 x^2 + 2 x - 1.
$$
answered 10 hours ago
Andreas CarantiAndreas Caranti
57k34397
$endgroup$
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,
$$
frac11 + omega = - omega,
quadtextandquad
frac11 + omega^* = - omega^*
$$
are the roots of $x^2 - x + 1$, and so
$$
1, - omega, - omega^*
$$
will be the roots of
$$
(x - 1) (x^2 - x + 1)
=
x^3 - 2 x^2 + 2 x - 1.
$$
answered 10 hours ago
Andreas CarantiAndreas Caranti
57k34397
answered 10 hours ago
Andreas CarantiAndreas Caranti
57k34397
answered 10 hours ago
Andreas CarantiAndreas Caranti
57k34397
answered 10 hours ago
Andreas CarantiAndreas Caranti
57k34397
57k34397
add a comment |
add a comment |
$begingroup$
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 $$z^3+az^2+bz+cequiv(z-1)left(z-frac11+omegaright)left(z-frac11+omega^*right)$$ To determine $a,b,c$, simply multiply this out and equate the coefficients of different powers of $z$.
answered 10 hours ago
John DoeJohn Doe
11.8k11239
$endgroup$
add a comment |
$begingroup$
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 $$z^3+az^2+bz+cequiv(z-1)left(z-frac11+omegaright)left(z-frac11+omega^*right)$$ To determine $a,b,c$, simply multiply this out and equate the coefficients of different powers of $z$.
answered 10 hours ago
John DoeJohn Doe
11.8k11239
$endgroup$
add a comment |
$begingroup$
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 $$z^3+az^2+bz+cequiv(z-1)left(z-frac11+omegaright)left(z-frac11+omega^*right)$$ To determine $a,b,c$, simply multiply this out and equate the coefficients of different powers of $z$.
answered 10 hours ago
John DoeJohn Doe
11.8k11239
$endgroup$
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 $$z^3+az^2+bz+cequiv(z-1)left(z-frac11+omegaright)left(z-frac11+omega^*right)$$ To determine $a,b,c$, simply multiply this out and equate the coefficients of different powers of $z$.
answered 10 hours ago
John DoeJohn Doe
11.8k11239
answered 10 hours ago
John DoeJohn Doe
11.8k11239
answered 10 hours ago
John DoeJohn Doe
11.8k11239
answered 10 hours ago
John DoeJohn Doe
11.8k11239
11.8k11239
add a comment |
add a comment |
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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
9 hours ago