MHB Solve Polynomial Equation z^6-2z^3+2=0

[Petrus]

Hello MHB,
Find all roots to $$z^6-2z^3+2=0$$
I can se we there will be 6 roots.
I start with subsitute $$z^3=t$$ so we got
$$t^2-2t+2=0$$ and we get $$t_1=1+i$$ and $$t_2=1-i$$
what shall I do next? Shall I go to polar form?
Regards,

 

[MarkFL]

Yes, that would be a good way to continue and find your six roots.

 

[Petrus]

Yes, that would be a good way to continue and find your six roots.

in that t I get $$\sqrt{2}e^{i \frac{\pi}{4}}$$ so I shall solve $$t^3=\sqrt{2}e^{i \frac{\pi}{4}}$$

 

[MarkFL]

Yes, in fact you could use:

$$t^3=\sqrt{2}e^{\pm\frac{\pi}{4}i}$$

to get all six roots.

 

[Petrus]

Yes, in fact you could use:

$$t^3=\sqrt{2}e^{\pm\frac{\pi}{4}i}$$

to get all six roots.

I still don't get six roots, I get $$2^{1/6}e^{\pm\frac{\pi}{12}i}$$ but that is not six roots.

 

[MarkFL]

If it were me, I would use Euler's formula to express $t^3$ in trigonometric form, then use de Moivre's theorem to get all six roots.

Or equivalently write:

$$t^3=\sqrt{2}e^{\pm\left(\frac{\pi}{4}+2k\pi \right)i}$$

Then let $$k\in\{-1,0,1\}$$ to get $$\theta\in[-\pi,\pi]$$

 

[Petrus]

If it were me, I would use Euler's formula to express $t^3$ in trigonometric form, then use de Moivre's theorem to get all six roots.

Is this correct?
$$2^{1/6}e^{\pm\frac{\pi}{12}+ \frac{k2\pi}{12}}$$ $$k=0,1,2,3,4,5$$

 

[MarkFL]

Is this correct?
$$2^{1/6}e^{\pm\frac{\pi}{12}+ \frac{k2\pi}{12}}$$ $$k=0,1,2,3,4,5$$

No, you have omitted $i$ from the exponent, and you want to get 3 roots from both values of $t^3$ not 6 roots from one value.

Please note I have added to my previous post on how to do this.

 

[Petrus]

No, you have omitted $i$ from the exponent, and you want to get 3 roots from both values of $t^3$ not 6 roots from one value.

Please note I have added to my previous post on how to do this.

Hello Mark,
I don't understand that one cause I can't find it on my book. The only one I could find is de moivres sats so we can write $$t=2^{1/6}e^{\pm( \frac{\pi}{12}+ \frac{k2\pi}{12})i}$$ $$k=0,1,2,3,4,5$$, I am pretty much clueless right now...

Regards,

 

[Petrus]

I start to understand what you do but now where you get your k and $$\theta$$

 

[Prove It]

Hello MHB,
Find all roots to $$z^6-2z^3+2=0$$
I can se we there will be 6 roots.
I start with subsitute $$z^3=t$$ so we got
$$t^2-2t+2=0$$ and we get $$t_1=1+i$$ and $$t_2=1-i$$
what shall I do next? Shall I go to polar form?
Regards,

It's always easiest to remember that there are as many solutions (at least in one rotation of a circle) to any complex polynomial as the degree of the polynomial, and they are all evenly spaced around a circle, so they all have the same magnitude and are separated by the same angle. So your sixth-degree polynomial has six solutions, and you have reduced it down to two cubics, each which will have three solutions.

For your first cubic, you have

[math]\displaystyle \begin{align*} z^3 &= 1 + i \\ z^3 &= \sqrt{2}\,e^{\frac{\pi}{4}\,i} \\ z &= \left( \sqrt{2} \, e^{\frac{\pi}{4}\,i} \right) ^{\frac{1}{3}} \\ z &= \sqrt[6]{2} \, e^{\frac{\pi}{12}\,i} \end{align*}[/math]

And now remembering that all three solutions have the same magnitude and are evenly spaced about a circle, they will be separated by an angle of [math]\displaystyle \begin{align*} \frac{2\pi}{3} \end{align*}[/math], and so the solutions are

[math]\displaystyle \begin{align*} z_1 &= \sqrt[6]{2} \, e^{\frac{\pi}{12}\,i} \\ \\ z_2 &= \sqrt[6]{2}\, e^{\frac{3\pi}{4}\,i} \\ \\ z_3 &= \sqrt[6]{2}\, e^{-\frac{7\pi}{12}\,i} \end{align*}[/math]Follow a similar process to find the solutions to [math]\displaystyle \begin{align*} z^3 = 1 - i \end{align*}[/math] :)

 

[Petrus]

Hello,
Now I do understand and get correct answer as facit!:) Thanks MarkFL and prove it for taking your time!

Regards,