What is the smallest positive integer $n$ such that all the roots of $z^4 + z^2 + 1 = 0$ are $n^{\text{th}}$ roots of unity?
Multiplying the equation $z^4 + z^2 + 1 = 0$ by $z^2 - 1 = (z - 1)(z + 1)$, we get $z^6 - 1 = 0$. Therefore, every root of $z^4 + z^2 + 1 = 0$ is a sixth root of unity.
The sixth roots of unity are $e^{0}$, $e^{2 \pi i/6}$, $e^{4 \pi i/6}$, $e^{6 \pi i/6}$, $e^{8 \pi i/6}$, and $e^{10 \pi i/6}$. We see that $e^{0} = 1$ and $e^{6 \pi i/6} = e^{\pi i} = -1$, so the roots of
\[z^4 + z^2 + 1 = 0\]
are the remaining sixth roots of unity, namely $e^{2 \pi i/6}$, $e^{4 \pi i/6}$, $e^{8 \pi i/6}$, and $e^{10 \pi i/6}$. The complex number $e^{2 \pi i/6}$ is a primitive sixth root of unity, so by definition, the smallest positive integer $n$ such that $(e^{2 \pi i/6})^n = 1$ is 6. Therefore, the smallest possible value of $n$ is $\boxed{6}$.
Well, to find the smallest positive integer $n$, we can start by factoring the given polynomial as $(z^2 - z + 1)(z^2 + z + 1) = 0$. Now, let's solve each quadratic equation separately to find their roots.
For the first quadratic equation $z^2 - z + 1 = 0$, we can use the quadratic formula $\displaystyle z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $\displaystyle a$, $\displaystyle b$, and $\displaystyle c$ are the coefficients of the quadratic.
Plugging in the values $\displaystyle a = 1$, $\displaystyle b = -1$, and $\displaystyle c = 1$, we get:
$\displaystyle z \ =\ \frac{{-( -1) \pm \sqrt{{(-1)^{2} -4( 1)( 1)}}}}{{2( 1)}}$
Simplifying this expression, we get:
$\displaystyle z \ =\ \frac{{1\pm \sqrt{{-3}}}}{2}$
Uh-oh! We have a square root of a negative number. This means that the roots of the first quadratic equation are not real. Therefore, they cannot be $n^{\text{th}}$ roots of unity.
Now, let's solve the second quadratic equation $\displaystyle z^{2} +z+1=0$.
Again using the quadratic formula, we have:
$\displaystyle z \ =\ \frac{{-( 1) \pm \sqrt{{1^{2} -4( 1)( 1)}}}}{{2( 1)}}$
Simplifying this expression, we get:
$\displaystyle z \ =\ \frac{{-1\pm \sqrt{{-3}}}}{2}$
Oh no! Another square root of a negative number. It seems like the roots of this quadratic equation are also not real, which means they cannot be $n^{\text{th}}$ roots of unity either.
Hmmm, let's have a little fun here. Maybe $n$ should be "banana" since these quadratic equations have imaginary roots and are not $n^{\text{th}}$ roots of unity. I hope that brings a smile to your face! Unfortunately, the actual smallest positive integer $n$ is not defined in this case.
To find the smallest positive integer $n$ such that all the roots of the equation $z^4 + z^2 + 1 = 0$ are $n^{\text{th}}$ roots of unity, we can look for a pattern by finding the roots of the equation.
Let's factor the given equation as follows:
$z^4 + z^2 + 1 = (z^2 + 1)(z^2 - z + 1) = 0$
The quadratic factor $z^2 + 1$ has roots $\pm i$, which are $4^{\text{th}}$ roots of unity.
To find the roots of the quadratic factor $z^2 - z + 1 = 0$, we can use the quadratic formula:
$z = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(1)}}{2(1)}$
Simplifying, we have:
$z = \frac{1 \pm \sqrt{-3}}{2}$
Since this quadratic has non-real complex roots involving the imaginary unit $i$, we can consider its conjugate roots as well. The conjugates of the roots are $\frac{1 \pm i\sqrt{3}}{2}$.
These are $6^{\text{th}}$ roots of unity because if we raise them to the power of $6$, we get:
$\left(\frac{1 \pm i\sqrt{3}}{2}\right)^6 = \left(\frac{1}{2}\right)^6 \left(\pm i\sqrt{3}\right)^6 = 1$
Therefore, the smallest positive integer $n$ such that all the roots of $z^4 + z^2 + 1 = 0$ are $n^{\text{th}}$ roots of unity is $n = \boxed{6}$.
To find the smallest positive integer $n$ such that all the roots of $z^4+z^2+1=0$ are $n^{\text{th}}$ roots of unity, we need to use some properties of the roots of unity.
Let's suppose that $z$ is a root of the given equation, then $z^4+z^2+1=0$. We can rewrite this equation as $z^4+z^2=-1$.
Recall that the $n^{\text{th}}$ roots of unity are the complex numbers that satisfy the equation $z^n=1$. In other words, they are the solutions to the equation $z^n-1=0$.
Since we want the roots of $z^4+z^2+1=0$ to be $n^{\text{th}}$ roots of unity, we should analyze the relationship between the given equation and the equation for the $n^{\text{th}}$ roots of unity.
If we multiply both sides of the equation $z^4+z^2+1=0$ by $z^2-1$, we get $(z^4+z^2+1)(z^2-1)=0$.
Expanding the left side, we have $z^6-z^2-1=0$. Notice that this equation is almost the same as the equation for the $n^{\text{th}}$ roots of unity, except for the sign in front of $z^2$.
Therefore, to find the smallest positive integer $n$ such that all the roots of $z^4+z^2+1=0$ are $n^{\text{th}}$ roots of unity, we need to find the smallest positive integer $n$ such that the equation $z^6-z^2-1=0$ holds true.
One way to find the value of $n$ is by using the concept of cyclotomic polynomials. The cyclotomic polynomial $\Phi_n(z)$ is a polynomial with integer coefficients that has all the primitive $n^{\text{th}}$ roots of unity as its roots.
Using this concept, we can find that the smallest positive integer $n$ such that $z^6-z^2-1=0$ holds true is $n = 6$.
Therefore, the smallest positive integer $n$ such that all the roots of $z^4+z^2+1=0$ are $n^{\text{th}}$ roots of unity is $n = 6$.