# de Moivre's Theorem

• May 5th 2011, 08:01 AM
domenfrandolic
de Moivre's Theorem
Hello.

when you see the\theta which is unknown

i need to solve the following by Moivres theorem

r^3(cos 3\theta + i sin 3\theta )=1..how can i find the \theta ?
i thought that the Real part equals 1 as you can see...and the imaginary part is 0 since is not there...so
r^3 cos3\theta =1
r^3 sin3\theta =0

so to find the angle \theta i tried r^3 sin3\theta devided by r^3 cos3\theta equals 0 devided by 1...then the next step would be sin3\theta devided cos3\theta equals 0...therefore tan3\theta =0...from now on i am stuck...but i am not sure it was the right step to follow...any advices how to get ahead?
• May 5th 2011, 08:13 AM
Plato
Quote:

Originally Posted by domenfrandolic
when you see the\theta which is unknown
i need to solve the following by Moivres theorem
r^3(cos 3\theta + i sin 3\theta )=1..how can i find the \theta ?

It seems as if you have just thrown us into the middle of a problem.
There is really no way to help you if you do not give us the original question in its whole form. Please state the entire question being solved.
• May 5th 2011, 10:14 AM
domenfrandolic
z^3-1=0...but it has to be solved by using the moivre's theorem
• May 5th 2011, 10:25 AM
Plato
Quote:

Originally Posted by domenfrandolic
z^3-1=0...but it has to be solved by using the moivre's theorem

Had you done that to begin with, would have gotten help much sooner.

$\displaystyle 1=\cos(0)+\mathbf{i}\sin(0)$. That is one cube root of 1.
The three cube roots are
$\displaystyle \cos\left(\frac{2k\pi}{3}\right)+\mathbf{i}\sin$$\displaystyle \left(\frac{2k\pi}{3}\right),~k=0,1,2 • May 6th 2011, 05:45 AM HallsofIvy Quote: Originally Posted by domenfrandolic Hello. when you see the\theta which is unknown i need to solve the following by Moivres theorem r^3(cos 3\theta + i sin 3\theta )=1..how can i find the \theta ? i thought that the Real part equals 1 as you can see...and the imaginary part is 0 since is not there...so r^3 cos3\theta =1 r^3 sin3\theta =0 so to find the angle \theta i tried r^3 sin3\theta devided by r^3 cos3\theta equals 0 devided by 1...then the next step would be sin3\theta devided cos3\theta equals 0...therefore tan3\theta =0...from now on i am stuck...but i am not sure it was the right step to follow...any advices how to get ahead? Since r^3 cos(3theta)= 1, r cannot be 0 and so from r^3 sin(theta)= 0, you have sin(theta)= 0. Either theta= 0 or theta= pi. If theta pi, then r^3 cos(theta)= -r^3= 1 which is impossible because r is never negative: theta= 0 so that r^3 cos(theta)= r^3= 1 and r= 1. That, is one solution to z^3= 1 is z= 1!!! If you found that difficult then finding the other two solutions is going to be really difficult. • May 8th 2011, 01:53 AM domenfrandolic Quote: Originally Posted by Plato Had you done that to begin with, would have gotten help much sooner. \displaystyle 1=\cos(0)+\mathbf{i}\sin(0). That is one cube root of 1. The three cube roots are \displaystyle \cos\left(\frac{2k\pi}{3}\right)+\mathbf{i}\sin$$\displaystyle \left(\frac{2k\pi}{3}\right),~k=0,1,2$

how did you come out for those cube roots?it looks very good though...so with it then i find 3 solutions right?
• May 8th 2011, 01:57 AM
domenfrandolic
Quote:

Originally Posted by HallsofIvy
Since r^3 cos(3theta)= 1, r cannot be 0 and so from r^3 sin(theta)= 0, you have sin(theta)= 0. Either theta= 0 or theta= pi. If theta pi, then r^3 cos(theta)= -r^3= 1 which is impossible because r is never negative: theta= 0 so that r^3 cos(theta)= r^3= 1 and r= 1. That, is one solution to z^3= 1 is z= 1!!! If you found that difficult then finding the other two solutions is going to be really difficult.

so basically yeah...so why you put that theta is either 0 or pi?for the rest is clear
• May 8th 2011, 03:28 AM
tonio
Quote:

Originally Posted by domenfrandolic
how did you come out for those cube roots?it looks very good though...so with it then i find 3 solutions right?

Writing $\displaystyle z^3 = r(\cos(\theta+2k\pi)+i\sin(\theta+2k\pi)=1$, you get by de Moivre's Formula

that $\displaystyle z = r^{1/3}\left(\cos\left(\frac{\theta+2k\pi}{3}\right)+i\ sin\left(\frac{\theta+2k\pi}{3}\right)\right)=1 , k=0,1,2$

(why can we take only those three integer values of k?) , and from here that the answer you've been given.

Tonio