Need help on understanding the solution (Linear Algebra done right)
$begingroup$
In exercise 1A Q2, the question wants us to show that $dfrac{-1+sqrt{3}i}{2} $ is a cube root of 1 (meaning that its cube equals 1). I can solve this by directly compute the answer. But I found another solution online which I don't quite understand.
It said:
Note that:
$$(a+bi)+(a-bi)=2a$$ , and
$$(a+bi)(a-bi)=a^2+b^2$$
It follows that $dfrac{-1+sqrt{3}i}{2} $ is the root of $x^2+x+1=0$
For, $$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1$$
and
$$frac{-1+sqrt{3}i}{2}frac{-1-sqrt{3}i}{2}=1.$$
Because $x^3-1=(x-1)(x^2+x+1)$, we obtain the solution.
I don't quite follow the logic. Can anyone help?
linear-algebra
asked Dec 1 '18 at 8:18
JOHN JOHN
1448
$endgroup$
add a comment |
$begingroup$
In exercise 1A Q2, the question wants us to show that $dfrac{-1+sqrt{3}i}{2} $ is a cube root of 1 (meaning that its cube equals 1). I can solve this by directly compute the answer. But I found another solution online which I don't quite understand.
It said:
Note that:
$$(a+bi)+(a-bi)=2a$$ , and
$$(a+bi)(a-bi)=a^2+b^2$$
It follows that $dfrac{-1+sqrt{3}i}{2} $ is the root of $x^2+x+1=0$
For, $$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1$$
and
$$frac{-1+sqrt{3}i}{2}frac{-1-sqrt{3}i}{2}=1.$$
Because $x^3-1=(x-1)(x^2+x+1)$, we obtain the solution.
I don't quite follow the logic. Can anyone help?
linear-algebra
asked Dec 1 '18 at 8:18
JOHN JOHN
1448
$endgroup$
add a comment |
$begingroup$
In exercise 1A Q2, the question wants us to show that $dfrac{-1+sqrt{3}i}{2} $ is a cube root of 1 (meaning that its cube equals 1). I can solve this by directly compute the answer. But I found another solution online which I don't quite understand.
It said:
Note that:
$$(a+bi)+(a-bi)=2a$$ , and
$$(a+bi)(a-bi)=a^2+b^2$$
It follows that $dfrac{-1+sqrt{3}i}{2} $ is the root of $x^2+x+1=0$
For, $$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1$$
and
$$frac{-1+sqrt{3}i}{2}frac{-1-sqrt{3}i}{2}=1.$$
Because $x^3-1=(x-1)(x^2+x+1)$, we obtain the solution.
I don't quite follow the logic. Can anyone help?
linear-algebra
asked Dec 1 '18 at 8:18
JOHN JOHN
1448
$endgroup$
In exercise 1A Q2, the question wants us to show that $dfrac{-1+sqrt{3}i}{2} $ is a cube root of 1 (meaning that its cube equals 1). I can solve this by directly compute the answer. But I found another solution online which I don't quite understand.
It said:
Note that:
$$(a+bi)+(a-bi)=2a$$ , and
$$(a+bi)(a-bi)=a^2+b^2$$
It follows that $dfrac{-1+sqrt{3}i}{2} $ is the root of $x^2+x+1=0$
For, $$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1$$
and
$$frac{-1+sqrt{3}i}{2}frac{-1-sqrt{3}i}{2}=1.$$
Because $x^3-1=(x-1)(x^2+x+1)$, we obtain the solution.
I don't quite follow the logic. Can anyone help?
linear-algebra
linear-algebra
asked Dec 1 '18 at 8:18
JOHN JOHN
1448
asked Dec 1 '18 at 8:18
JOHN JOHN
1448
asked Dec 1 '18 at 8:18
JOHN JOHN
1448
asked Dec 1 '18 at 8:18
JOHN JOHN
1448
asked Dec 1 '18 at 8:18
JOHN JOHN
1448
1448
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
If $x_1$ and $x_2$ are the roots of $x^2+x+1$, then
$$ x^2+x+1 = (x-x_1)(x-x_2) = x^2 -(x_1+x_2)x + x_1x_2.$$
This implies that $x_1,x_2$ are the unique numbers satisfying $$ x_1+x_2=-1, x_1x_2 = 1. $$
So $x_1,x_2=(-1 pm sqrt 3)/2$ are the roots of $x^2+x+1$, since they satisfy the above equations. They are also roots of $x^3-1$ (ie, cube roots of 1) due to the factorization you wrote in the second last line.
answered Dec 1 '18 at 8:47
QuaternionQuaternion
1966
$endgroup$
add a comment |
$begingroup$
If $z_1$ and $z_2$ are roots of
$$(z-z_1)(z-z_2)=0$$
$$z^2-(z_1+z_2)z+z_1z_2=0$$
In particular, $z_1$ and $bar{z_1}$ are roots of
$$z^2-2Re(z_1)z+|z_1|^2=0$$
Hence, we can see that $frac{-1+sqrt3i}{2}$ is a root of $$x^2+x+1=0$$
Hence it is also a root of $$x^3-1=(x-1)(x^2+x+1).$$
answered Dec 1 '18 at 8:26
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
The solution is trying to form a quadratic equation one of whose roots is $frac{-1+isqrt3}2$. Recall that for a quadratic equation with real coefficients, complex roots always occur in conjugate pairs. This means that the other root of the quadratic must be $frac{-1-isqrt3}2$. The remaining procedure illustrates how to quickly guess the quadratic given its roots.
Note that for the quadratic equation $ax^2+bx+c=0$, the sum of roots is equal to $-b/a$ and their product $c/a.$
$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1=-b/a, frac{-1+sqrt{3}i}{2}cdotfrac{-1-sqrt{3}i}{2}=1=c/aimplies a=c=b=1$
The required quadratic equation is $x^2+x+1=0implies x^2=-x-1implies x^3=-x^2-x=1$
answered Dec 1 '18 at 9:37
Shubham JohriShubham Johri
4,594717
$endgroup$
add a comment |
$begingroup$
If $frac{-1+isqrt{3}} {2} $ is a square root of 1 it means that it's the solution of $x^3-1=0$ by definition. You can split it in $(x-1)(x^2+x+1)=0$ and so you have all the cubic solution, $1$ and the other two found with the splitting of the second polynomial with the sum and difference method.
answered Dec 1 '18 at 8:25
TheoopTheoop
1
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $x_1$ and $x_2$ are the roots of $x^2+x+1$, then
$$ x^2+x+1 = (x-x_1)(x-x_2) = x^2 -(x_1+x_2)x + x_1x_2.$$
This implies that $x_1,x_2$ are the unique numbers satisfying $$ x_1+x_2=-1, x_1x_2 = 1. $$
So $x_1,x_2=(-1 pm sqrt 3)/2$ are the roots of $x^2+x+1$, since they satisfy the above equations. They are also roots of $x^3-1$ (ie, cube roots of 1) due to the factorization you wrote in the second last line.
answered Dec 1 '18 at 8:47
QuaternionQuaternion
1966
$endgroup$
add a comment |
$begingroup$
If $x_1$ and $x_2$ are the roots of $x^2+x+1$, then
$$ x^2+x+1 = (x-x_1)(x-x_2) = x^2 -(x_1+x_2)x + x_1x_2.$$
This implies that $x_1,x_2$ are the unique numbers satisfying $$ x_1+x_2=-1, x_1x_2 = 1. $$
So $x_1,x_2=(-1 pm sqrt 3)/2$ are the roots of $x^2+x+1$, since they satisfy the above equations. They are also roots of $x^3-1$ (ie, cube roots of 1) due to the factorization you wrote in the second last line.
answered Dec 1 '18 at 8:47
QuaternionQuaternion
1966
$endgroup$
add a comment |
$begingroup$
If $x_1$ and $x_2$ are the roots of $x^2+x+1$, then
$$ x^2+x+1 = (x-x_1)(x-x_2) = x^2 -(x_1+x_2)x + x_1x_2.$$
This implies that $x_1,x_2$ are the unique numbers satisfying $$ x_1+x_2=-1, x_1x_2 = 1. $$
So $x_1,x_2=(-1 pm sqrt 3)/2$ are the roots of $x^2+x+1$, since they satisfy the above equations. They are also roots of $x^3-1$ (ie, cube roots of 1) due to the factorization you wrote in the second last line.
answered Dec 1 '18 at 8:47
QuaternionQuaternion
1966
$endgroup$
If $x_1$ and $x_2$ are the roots of $x^2+x+1$, then
$$ x^2+x+1 = (x-x_1)(x-x_2) = x^2 -(x_1+x_2)x + x_1x_2.$$
This implies that $x_1,x_2$ are the unique numbers satisfying $$ x_1+x_2=-1, x_1x_2 = 1. $$
So $x_1,x_2=(-1 pm sqrt 3)/2$ are the roots of $x^2+x+1$, since they satisfy the above equations. They are also roots of $x^3-1$ (ie, cube roots of 1) due to the factorization you wrote in the second last line.
answered Dec 1 '18 at 8:47
QuaternionQuaternion
1966
answered Dec 1 '18 at 8:47
QuaternionQuaternion
1966
answered Dec 1 '18 at 8:47
QuaternionQuaternion
1966
answered Dec 1 '18 at 8:47
QuaternionQuaternion
1966
1966
add a comment |
add a comment |
$begingroup$
If $z_1$ and $z_2$ are roots of
$$(z-z_1)(z-z_2)=0$$
$$z^2-(z_1+z_2)z+z_1z_2=0$$
In particular, $z_1$ and $bar{z_1}$ are roots of
$$z^2-2Re(z_1)z+|z_1|^2=0$$
Hence, we can see that $frac{-1+sqrt3i}{2}$ is a root of $$x^2+x+1=0$$
Hence it is also a root of $$x^3-1=(x-1)(x^2+x+1).$$
answered Dec 1 '18 at 8:26
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
If $z_1$ and $z_2$ are roots of
$$(z-z_1)(z-z_2)=0$$
$$z^2-(z_1+z_2)z+z_1z_2=0$$
In particular, $z_1$ and $bar{z_1}$ are roots of
$$z^2-2Re(z_1)z+|z_1|^2=0$$
Hence, we can see that $frac{-1+sqrt3i}{2}$ is a root of $$x^2+x+1=0$$
Hence it is also a root of $$x^3-1=(x-1)(x^2+x+1).$$
answered Dec 1 '18 at 8:26
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
If $z_1$ and $z_2$ are roots of
$$(z-z_1)(z-z_2)=0$$
$$z^2-(z_1+z_2)z+z_1z_2=0$$
In particular, $z_1$ and $bar{z_1}$ are roots of
$$z^2-2Re(z_1)z+|z_1|^2=0$$
Hence, we can see that $frac{-1+sqrt3i}{2}$ is a root of $$x^2+x+1=0$$
Hence it is also a root of $$x^3-1=(x-1)(x^2+x+1).$$
answered Dec 1 '18 at 8:26
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
If $z_1$ and $z_2$ are roots of
$$(z-z_1)(z-z_2)=0$$
$$z^2-(z_1+z_2)z+z_1z_2=0$$
In particular, $z_1$ and $bar{z_1}$ are roots of
$$z^2-2Re(z_1)z+|z_1|^2=0$$
Hence, we can see that $frac{-1+sqrt3i}{2}$ is a root of $$x^2+x+1=0$$
Hence it is also a root of $$x^3-1=(x-1)(x^2+x+1).$$
answered Dec 1 '18 at 8:26
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 1 '18 at 8:26
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 1 '18 at 8:26
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 1 '18 at 8:26
Siong Thye GohSiong Thye Goh
100k1465117
100k1465117
add a comment |
add a comment |
$begingroup$
The solution is trying to form a quadratic equation one of whose roots is $frac{-1+isqrt3}2$. Recall that for a quadratic equation with real coefficients, complex roots always occur in conjugate pairs. This means that the other root of the quadratic must be $frac{-1-isqrt3}2$. The remaining procedure illustrates how to quickly guess the quadratic given its roots.
Note that for the quadratic equation $ax^2+bx+c=0$, the sum of roots is equal to $-b/a$ and their product $c/a.$
$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1=-b/a, frac{-1+sqrt{3}i}{2}cdotfrac{-1-sqrt{3}i}{2}=1=c/aimplies a=c=b=1$
The required quadratic equation is $x^2+x+1=0implies x^2=-x-1implies x^3=-x^2-x=1$
answered Dec 1 '18 at 9:37
Shubham JohriShubham Johri
4,594717
$endgroup$
add a comment |
$begingroup$
The solution is trying to form a quadratic equation one of whose roots is $frac{-1+isqrt3}2$. Recall that for a quadratic equation with real coefficients, complex roots always occur in conjugate pairs. This means that the other root of the quadratic must be $frac{-1-isqrt3}2$. The remaining procedure illustrates how to quickly guess the quadratic given its roots.
Note that for the quadratic equation $ax^2+bx+c=0$, the sum of roots is equal to $-b/a$ and their product $c/a.$
$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1=-b/a, frac{-1+sqrt{3}i}{2}cdotfrac{-1-sqrt{3}i}{2}=1=c/aimplies a=c=b=1$
The required quadratic equation is $x^2+x+1=0implies x^2=-x-1implies x^3=-x^2-x=1$
answered Dec 1 '18 at 9:37
Shubham JohriShubham Johri
4,594717
$endgroup$
add a comment |
$begingroup$
The solution is trying to form a quadratic equation one of whose roots is $frac{-1+isqrt3}2$. Recall that for a quadratic equation with real coefficients, complex roots always occur in conjugate pairs. This means that the other root of the quadratic must be $frac{-1-isqrt3}2$. The remaining procedure illustrates how to quickly guess the quadratic given its roots.
Note that for the quadratic equation $ax^2+bx+c=0$, the sum of roots is equal to $-b/a$ and their product $c/a.$
$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1=-b/a, frac{-1+sqrt{3}i}{2}cdotfrac{-1-sqrt{3}i}{2}=1=c/aimplies a=c=b=1$
The required quadratic equation is $x^2+x+1=0implies x^2=-x-1implies x^3=-x^2-x=1$
answered Dec 1 '18 at 9:37
Shubham JohriShubham Johri
4,594717
$endgroup$
The solution is trying to form a quadratic equation one of whose roots is $frac{-1+isqrt3}2$. Recall that for a quadratic equation with real coefficients, complex roots always occur in conjugate pairs. This means that the other root of the quadratic must be $frac{-1-isqrt3}2$. The remaining procedure illustrates how to quickly guess the quadratic given its roots.
Note that for the quadratic equation $ax^2+bx+c=0$, the sum of roots is equal to $-b/a$ and their product $c/a.$
$frac{-1+sqrt{3}i}{2}+frac{-1-sqrt{3}i}{2}=-1=-b/a, frac{-1+sqrt{3}i}{2}cdotfrac{-1-sqrt{3}i}{2}=1=c/aimplies a=c=b=1$
The required quadratic equation is $x^2+x+1=0implies x^2=-x-1implies x^3=-x^2-x=1$
answered Dec 1 '18 at 9:37
Shubham JohriShubham Johri
4,594717
answered Dec 1 '18 at 9:37
Shubham JohriShubham Johri
4,594717
answered Dec 1 '18 at 9:37
Shubham JohriShubham Johri
4,594717
answered Dec 1 '18 at 9:37
Shubham JohriShubham Johri
4,594717
4,594717
add a comment |
add a comment |
$begingroup$
If $frac{-1+isqrt{3}} {2} $ is a square root of 1 it means that it's the solution of $x^3-1=0$ by definition. You can split it in $(x-1)(x^2+x+1)=0$ and so you have all the cubic solution, $1$ and the other two found with the splitting of the second polynomial with the sum and difference method.
answered Dec 1 '18 at 8:25
TheoopTheoop
1
$endgroup$
add a comment |
$begingroup$
If $frac{-1+isqrt{3}} {2} $ is a square root of 1 it means that it's the solution of $x^3-1=0$ by definition. You can split it in $(x-1)(x^2+x+1)=0$ and so you have all the cubic solution, $1$ and the other two found with the splitting of the second polynomial with the sum and difference method.
answered Dec 1 '18 at 8:25
TheoopTheoop
1
$endgroup$
add a comment |
$begingroup$
If $frac{-1+isqrt{3}} {2} $ is a square root of 1 it means that it's the solution of $x^3-1=0$ by definition. You can split it in $(x-1)(x^2+x+1)=0$ and so you have all the cubic solution, $1$ and the other two found with the splitting of the second polynomial with the sum and difference method.
answered Dec 1 '18 at 8:25
TheoopTheoop
1
$endgroup$
If $frac{-1+isqrt{3}} {2} $ is a square root of 1 it means that it's the solution of $x^3-1=0$ by definition. You can split it in $(x-1)(x^2+x+1)=0$ and so you have all the cubic solution, $1$ and the other two found with the splitting of the second polynomial with the sum and difference method.
answered Dec 1 '18 at 8:25
TheoopTheoop
1
answered Dec 1 '18 at 8:25
TheoopTheoop
1
answered Dec 1 '18 at 8:25
TheoopTheoop
1
answered Dec 1 '18 at 8:25
TheoopTheoop
1
1
add a comment |
add a comment |
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