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Complex number

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
Find all roots to \(\displaystyle z^6-2z^3+2=0\)
I can se we there will be 6 roots.
I start with subsitute \(\displaystyle z^3=t\) so we got
\(\displaystyle t^2-2t+2=0\) and we get \(\displaystyle t_1=1+i\) and \(\displaystyle t_2=1-i\)
what shall I do next? Shall I go to polar form?
Regards,
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, that would be a good way to continue and find your six roots.
 

Petrus

Well-known member
Feb 21, 2013
739
Yes, that would be a good way to continue and find your six roots.
in that t I get \(\displaystyle \sqrt{2}e^{i \frac{\pi}{4}}\) so I shall solve \(\displaystyle t^3=\sqrt{2}e^{i \frac{\pi}{4}}\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, in fact you could use:

\(\displaystyle t^3=\sqrt{2}e^{\pm\frac{\pi}{4}i}\)

to get all six roots.
 

Petrus

Well-known member
Feb 21, 2013
739
Yes, in fact you could use:

\(\displaystyle t^3=\sqrt{2}e^{\pm\frac{\pi}{4}i}\)

to get all six roots.
I still dont get six roots, I get \(\displaystyle 2^{1/6}e^{\pm\frac{\pi}{12}i}\) but that is not six roots.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

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

Then let \(\displaystyle k\in\{-1,0,1\}\) to get \(\displaystyle \theta\in[-\pi,\pi]\)
 
Last edited:

Petrus

Well-known member
Feb 21, 2013
739
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?
\(\displaystyle 2^{1/6}e^{\pm\frac{\pi}{12}+ \frac{k2\pi}{12}}\) \(\displaystyle k=0,1,2,3,4,5\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Is this correct?
\(\displaystyle 2^{1/6}e^{\pm\frac{\pi}{12}+ \frac{k2\pi}{12}}\) \(\displaystyle k=0,1,2,3,4,5\)
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

Well-known member
Feb 21, 2013
739
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 \(\displaystyle t=2^{1/6}e^{\pm( \frac{\pi}{12}+ \frac{k2\pi}{12})i}\) \(\displaystyle k=0,1,2,3,4,5\), I am pretty much clueless right now...

Regards,
 

Petrus

Well-known member
Feb 21, 2013
739
I start to understand what you do but now where you get your k and \(\displaystyle \theta\)
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Hello MHB,
Find all roots to \(\displaystyle z^6-2z^3+2=0\)
I can se we there will be 6 roots.
I start with subsitute \(\displaystyle z^3=t\) so we got
\(\displaystyle t^2-2t+2=0\) and we get \(\displaystyle t_1=1+i\) and \(\displaystyle t_2=1-i\)
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

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

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 \(\displaystyle \displaystyle \begin{align*} \frac{2\pi}{3} \end{align*}\), and so the solutions are

\(\displaystyle \displaystyle \begin{align*} z_1 &= \sqrt[6]{2} \, e^{\frac{\pi}{12}\,i} \\ \\ z_2 &= \sqrt[6]{2}\, e^{\frac{3\pi}{4}\,i} \\ \\ z_3 &= \sqrt[6]{2}\, e^{-\frac{7\pi}{12}\,i} \end{align*}\)


Follow a similar process to find the solutions to \(\displaystyle \displaystyle \begin{align*} z^3 = 1 - i \end{align*}\) :)
 

Petrus

Well-known member
Feb 21, 2013
739
Hello,
Now I do understand and get correct answer as facit!:) Thanks MarkFL and prove it for taking your time!

Regards,