Mixing
Why is $w^2$ a complex root for the equation $x^2+x+1=0$?
$begingroup$
"Multiplying both sides by $(x-1)$ gives $x^3-1=0$ which is the 3rd roots of unity. Its complex roots are also denoted as $w$ and $w^2$."
complex-numbers
asked Jan 19 at 20:53
N.CN.C
61
$endgroup$
add a comment |
$begingroup$
"Multiplying both sides by $(x-1)$ gives $x^3-1=0$ which is the 3rd roots of unity. Its complex roots are also denoted as $w$ and $w^2$."
complex-numbers
asked Jan 19 at 20:53
N.CN.C
61
$endgroup$
1
$begingroup$
Please write precisely what you are asking for.
$endgroup$
– szw1710
Jan 19 at 20:54
2
$begingroup$
You can just use the quadratic formula on the original equation. Are you asking why we call this $w$? Often it is $omega$. It is just a number that comes up somewhat frequently, so we give it a name.
$endgroup$
– Ross Millikan
Jan 19 at 20:56
$begingroup$
You have to understand what's $w$. Suppose $w$ is a cubic root of the unity, so in particular $|w|=1$. Then, multiplying by $overline{w}$, in the equation $w^3=1$ you get $w^2=overline{w}$. And you know that non-real roots of polynomials with real coefficients are there by conjugate pairs. In this way $w^2$ must be a root since it's the conjugate of a root. You can use a different reasoning to prove every power of a root of unity is a root of unity, too.
$endgroup$
– David Molano
Jan 19 at 20:59
add a comment |
$begingroup$
"Multiplying both sides by $(x-1)$ gives $x^3-1=0$ which is the 3rd roots of unity. Its complex roots are also denoted as $w$ and $w^2$."
complex-numbers
asked Jan 19 at 20:53
N.CN.C
61
$endgroup$
"Multiplying both sides by $(x-1)$ gives $x^3-1=0$ which is the 3rd roots of unity. Its complex roots are also denoted as $w$ and $w^2$."
complex-numbers
complex-numbers
asked Jan 19 at 20:53
N.CN.C
61
asked Jan 19 at 20:53
N.CN.C
61
asked Jan 19 at 20:53
N.CN.C
61
asked Jan 19 at 20:53
N.CN.C
61
asked Jan 19 at 20:53
N.CN.C
61
61
1
$begingroup$
Please write precisely what you are asking for.
$endgroup$
– szw1710
Jan 19 at 20:54
2
$begingroup$
You can just use the quadratic formula on the original equation. Are you asking why we call this $w$? Often it is $omega$. It is just a number that comes up somewhat frequently, so we give it a name.
$endgroup$
– Ross Millikan
Jan 19 at 20:56
$begingroup$
You have to understand what's $w$. Suppose $w$ is a cubic root of the unity, so in particular $|w|=1$. Then, multiplying by $overline{w}$, in the equation $w^3=1$ you get $w^2=overline{w}$. And you know that non-real roots of polynomials with real coefficients are there by conjugate pairs. In this way $w^2$ must be a root since it's the conjugate of a root. You can use a different reasoning to prove every power of a root of unity is a root of unity, too.
$endgroup$
– David Molano
Jan 19 at 20:59
add a comment |
1
$begingroup$
Please write precisely what you are asking for.
$endgroup$
– szw1710
Jan 19 at 20:54
2
$begingroup$
You can just use the quadratic formula on the original equation. Are you asking why we call this $w$? Often it is $omega$. It is just a number that comes up somewhat frequently, so we give it a name.
$endgroup$
– Ross Millikan
Jan 19 at 20:56
$begingroup$
You have to understand what's $w$. Suppose $w$ is a cubic root of the unity, so in particular $|w|=1$. Then, multiplying by $overline{w}$, in the equation $w^3=1$ you get $w^2=overline{w}$. And you know that non-real roots of polynomials with real coefficients are there by conjugate pairs. In this way $w^2$ must be a root since it's the conjugate of a root. You can use a different reasoning to prove every power of a root of unity is a root of unity, too.
$endgroup$
– David Molano
Jan 19 at 20:59
1
1
$begingroup$
Please write precisely what you are asking for.
$endgroup$
– szw1710
Jan 19 at 20:54
$begingroup$
Please write precisely what you are asking for.
$endgroup$
– szw1710
Jan 19 at 20:54
2
2
$begingroup$
You can just use the quadratic formula on the original equation. Are you asking why we call this $w$? Often it is $omega$. It is just a number that comes up somewhat frequently, so we give it a name.
$endgroup$
– Ross Millikan
Jan 19 at 20:56
$begingroup$
You can just use the quadratic formula on the original equation. Are you asking why we call this $w$? Often it is $omega$. It is just a number that comes up somewhat frequently, so we give it a name.
$endgroup$
– Ross Millikan
Jan 19 at 20:56
$begingroup$
You have to understand what's $w$. Suppose $w$ is a cubic root of the unity, so in particular $|w|=1$. Then, multiplying by $overline{w}$, in the equation $w^3=1$ you get $w^2=overline{w}$. And you know that non-real roots of polynomials with real coefficients are there by conjugate pairs. In this way $w^2$ must be a root since it's the conjugate of a root. You can use a different reasoning to prove every power of a root of unity is a root of unity, too.
$endgroup$
– David Molano
Jan 19 at 20:59
$begingroup$
You have to understand what's $w$. Suppose $w$ is a cubic root of the unity, so in particular $|w|=1$. Then, multiplying by $overline{w}$, in the equation $w^3=1$ you get $w^2=overline{w}$. And you know that non-real roots of polynomials with real coefficients are there by conjugate pairs. In this way $w^2$ must be a root since it's the conjugate of a root. You can use a different reasoning to prove every power of a root of unity is a root of unity, too.
$endgroup$
– David Molano
Jan 19 at 20:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $w$ be a complex root of $x^2+x+1$. Then $w^2+w+1=0$ so $w^3-1 = (w^2+w+1)(w-1) = 0$, so $w^3=1$, so $w^4=w$. Therefore $w^4+w^2+1=w+w^2+1=w^2+w+1=0$, so $w^2$ is also a root of $x^2+x+1$.
answered Jan 19 at 23:52
J. W. TannerJ. W. Tanner
2,6161217
$endgroup$
add a comment |
$begingroup$
Its pretty obvious after you examime the geometric sum;
$$x^2+x+1=frac{x^3-1}{x-1}$$
In general
$$sum_{j=0}^{n-1} x^j=frac{x^n-1}{x-1}$$
answered Jan 19 at 21:23
Andrew KovácsAndrew Kovács
1264
$endgroup$
add a comment |
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2 Answers
2
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2 Answers
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$begingroup$
Let $w$ be a complex root of $x^2+x+1$. Then $w^2+w+1=0$ so $w^3-1 = (w^2+w+1)(w-1) = 0$, so $w^3=1$, so $w^4=w$. Therefore $w^4+w^2+1=w+w^2+1=w^2+w+1=0$, so $w^2$ is also a root of $x^2+x+1$.
answered Jan 19 at 23:52
J. W. TannerJ. W. Tanner
2,6161217
$endgroup$
add a comment |
$begingroup$
Let $w$ be a complex root of $x^2+x+1$. Then $w^2+w+1=0$ so $w^3-1 = (w^2+w+1)(w-1) = 0$, so $w^3=1$, so $w^4=w$. Therefore $w^4+w^2+1=w+w^2+1=w^2+w+1=0$, so $w^2$ is also a root of $x^2+x+1$.
answered Jan 19 at 23:52
J. W. TannerJ. W. Tanner
2,6161217
$endgroup$
add a comment |
$begingroup$
Let $w$ be a complex root of $x^2+x+1$. Then $w^2+w+1=0$ so $w^3-1 = (w^2+w+1)(w-1) = 0$, so $w^3=1$, so $w^4=w$. Therefore $w^4+w^2+1=w+w^2+1=w^2+w+1=0$, so $w^2$ is also a root of $x^2+x+1$.
answered Jan 19 at 23:52
J. W. TannerJ. W. Tanner
2,6161217
$endgroup$
Let $w$ be a complex root of $x^2+x+1$. Then $w^2+w+1=0$ so $w^3-1 = (w^2+w+1)(w-1) = 0$, so $w^3=1$, so $w^4=w$. Therefore $w^4+w^2+1=w+w^2+1=w^2+w+1=0$, so $w^2$ is also a root of $x^2+x+1$.
answered Jan 19 at 23:52
J. W. TannerJ. W. Tanner
2,6161217
answered Jan 19 at 23:52
J. W. TannerJ. W. Tanner
2,6161217
answered Jan 19 at 23:52
J. W. TannerJ. W. Tanner
2,6161217
answered Jan 19 at 23:52
J. W. TannerJ. W. Tanner
2,6161217
2,6161217
add a comment |
add a comment |
$begingroup$
Its pretty obvious after you examime the geometric sum;
$$x^2+x+1=frac{x^3-1}{x-1}$$
In general
$$sum_{j=0}^{n-1} x^j=frac{x^n-1}{x-1}$$
answered Jan 19 at 21:23
Andrew KovácsAndrew Kovács
1264
$endgroup$
add a comment |
$begingroup$
Its pretty obvious after you examime the geometric sum;
$$x^2+x+1=frac{x^3-1}{x-1}$$
In general
$$sum_{j=0}^{n-1} x^j=frac{x^n-1}{x-1}$$
answered Jan 19 at 21:23
Andrew KovácsAndrew Kovács
1264
$endgroup$
add a comment |
$begingroup$
Its pretty obvious after you examime the geometric sum;
$$x^2+x+1=frac{x^3-1}{x-1}$$
In general
$$sum_{j=0}^{n-1} x^j=frac{x^n-1}{x-1}$$
answered Jan 19 at 21:23
Andrew KovácsAndrew Kovács
1264
$endgroup$
Its pretty obvious after you examime the geometric sum;
$$x^2+x+1=frac{x^3-1}{x-1}$$
In general
$$sum_{j=0}^{n-1} x^j=frac{x^n-1}{x-1}$$
answered Jan 19 at 21:23
Andrew KovácsAndrew Kovács
1264
answered Jan 19 at 21:23
Andrew KovácsAndrew Kovács
1264
answered Jan 19 at 21:23
Andrew KovácsAndrew Kovács
1264
answered Jan 19 at 21:23
Andrew KovácsAndrew Kovács
1264
1264
add a comment |
add a comment |
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1
$begingroup$
Please write precisely what you are asking for.
$endgroup$
– szw1710
Jan 19 at 20:54
2
$begingroup$
You can just use the quadratic formula on the original equation. Are you asking why we call this $w$? Often it is $omega$. It is just a number that comes up somewhat frequently, so we give it a name.
$endgroup$
– Ross Millikan
Jan 19 at 20:56
$begingroup$
You have to understand what's $w$. Suppose $w$ is a cubic root of the unity, so in particular $|w|=1$. Then, multiplying by $overline{w}$, in the equation $w^3=1$ you get $w^2=overline{w}$. And you know that non-real roots of polynomials with real coefficients are there by conjugate pairs. In this way $w^2$ must be a root since it's the conjugate of a root. You can use a different reasoning to prove every power of a root of unity is a root of unity, too.
$endgroup$
– David Molano
Jan 19 at 20:59