Solving complex polynomial equations

Question

I'm trying to solve this:

$3z^4 -z^3 +2z^2 -z +3 = 0$ "given no root is real"

The question gives us: $\frac {1} {z^n} +z^n = 2Cos(n \theta)$ and says hence or otherwise solve the above. I'm a bit stumped as i'm only getting real solutions.

I'm dividing through by $z^2$ to get: $3(z^2 +\frac {1}{z^2}) -(z+\frac {1}{z}) +2 = 0$, and then when I use the above identity I get:

$6Cos(2\theta) -2Cos(\theta) +2 = 0$. After using trig identity and solving: $6Cos^2(\theta) -Cos(\theta) -2 = 0$, I get real roots: $Cos(\theta) = \frac {2}{3}$, and $Cos(\theta) =-\frac {1}{2} $.

I'm not sure what i'm doing wrong. If someone could point me in the right direction, i'd be most grateful. Thanks.

Answer 1

Apart from a typo in your equation where the middle term should be $-2\cos(\theta)$ instead of $-\cos(\theta)$, I also got the same results as you for the values of $\cos(\theta)$.

As for what to do next, substitute these values in what the original question gives to determine $z$. For example, with $\cos(\theta) = \frac{2}{3}$, you get

$$\begin{equation}\begin{aligned} \frac{1}{z} + z & = 2\left(\frac{2}{3}\right) \\ 3 + 3z^2 & = 4z \\ 3z^2 - 4z + 3 & = 0 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Now, you can use the quadratic formula to get

$$z = \frac{4 \pm \sqrt{4^2 - 4(9)}}{6} = \frac{4 \pm 2\sqrt{-5}}{6} = \frac{2 \pm \sqrt{5}i}{3} \tag{2}\label{eq2A}$$

I'll leave it to you to handle the case where $\cos(\theta) = -\frac{1}{2}$ to get the other $2$ values of $z$.

Note that although the RHS of the formula you're given, i.e., $2\cos(n\theta)$, is real, it doesn't mean the $z$ on the LHS has to be real, as \eqref{eq2A} shows.

Answer 2

You are right the polynomial has no real roots

$3*z^4-z^3+2*z^2-z+3 = 0$

It's roots are $z_1 = -(\sqrt(3)*i+1)/2, z_2 = (\sqrt(3)*i-1)/2,z_3 = -(\sqrt(5)*i-2)/3,z_4 = (\sqrt(5)*i+2)/3$

But you made a mistake in your calculations, dividing the equation by $z^2$ to get $3*(z^2+1/(z^2))-(z+1/(z))+2 = 0$ $3×2*\cos(2×\theta)-2×\cos(\theta)+2 = 0$

$3×2×( 2×\cos(\theta)^2-1)-2×\cos(\theta)+2 = 0$ $3×2×2×\cos(\theta)^2-3×2-2×\cos(\theta)+2 = 0$ $12×\cos(\theta)^2-2×\cos(\theta)-4 = 0$

Now we have to solve this quadratic, which has roots $\cos(\theta) = -1/2$ and $\cos(\theta) = 2/3$ Then $\theta = \arccos(-1/2)$ and $\theta = \arccos(2/3)$

Now the angle is this ( working in radian )

$\theta = 2*\pi/3$ and $\theta = \arccos(2/3)$

I suggest you understand the true meaning of that identity and de' moivres theorem, because the $\theta$ contains the natural logarithm.

$\theta = -i×\log(z)$

Where $z$ is the root of your polynomial, therefore $z = exp(\theta × i )$

Which then becomes $z = exp((2*\pi/3)*i)$ and $z = exp(\arccos(2/3)*i)$

Which are the roots of your polynomial, offcourse it's conjugate is also a root