How do I completely solve the equation $z^4 - 2z^3 + 9z^2 - 14z + 14 = 0$ where there is a root with the real part of $1$.

Question

I would please like some help with solving the following equation:

$$z^4 - 2z^3 + 9z^2 - 14z + 14 = 0$$

All I know about the equation is that there is a root with the real part of $1$.

My approach has been to factor out the root: $1 + yi$ and divide the equation with it. The following calculations (using the division algorithm) is quite cumbersome, and I wonder if there is any better way of doing this?

Thank you kindly for your help!

Answer 1

As the coefficients of the different powers of $z$ are real,

if one of the four roots is $1+yi,$ the other must be its conjugate $1-yi$ .

If the other two roots are $x,w,$

then using Vieta's Formulas for the coefficient of $z^3$, $1+yi+1-yi+x+w=2\implies w=-x$

So, $$(z-x)(z+x)\{z-(1+yi)\}\{z-(1-yi)\}=0$$

$$\implies (z^2-x^2)(z^2-2z+1+y^2)=0$$

$$\implies z^4-2z^3+z^2(1+y^2-x^2)+2x^2z-x^2(1+y^2)=0$$

Now compare the coefficients of the different powers of $z$

So, $2x^2=-14,x^2=-7,x=\pm \sqrt{-7}=\pm \sqrt7i$ and $1+y^2-x^2=9\implies 1+y^2+7=9,y^2=1$

So, the roots are $\pm \sqrt7i,1\pm i$

Answer 2

Choosing $z = 1 + y i$, you obtain the formula:

$$y^4 - 2 i y^3 - 9y^2 + 2 i y + 8 = 0$$

Because $y$ is real, this means that

$$2 y^3 - 2y = 0$$

and

$$ y^4 - 9y^2 + 8 = 0$$

The first equation implies $y = 0, 1, -1$. The second equation implies that $y^2 = 8$ or $y^2 = 1$. Thus we see both $1 + i$ and $1 - i$ are solutions. Factor out the polynomial $(z - 1 + i)(z - 1 - i) = z^2 - 2z + 2$ and you'll find the remaining roots.

Answer 3

Maxima tells me the roots are $\pm i \sqrt{7}$, $1 \pm i$. It factors as $(z^2 + 7) (z^2 - 2 z + 2)$.