1. ## 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?

2. 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.

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

4. 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.

$1=\cos(0)+\mathbf{i}\sin(0)$. That is one cube root of 1.
The three cube roots are
$\cos\left(\frac{2k\pi}{3}\right)+\mathbf{i}\sin$ $\left(\frac{2k\pi}{3}\right),~k=0,1,2$

5. 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.

6. Originally Posted by Plato
Had you done that to begin with, would have gotten help much sooner.

$1=\cos(0)+\mathbf{i}\sin(0)$. That is one cube root of 1.
The three cube roots are
$\cos\left(\frac{2k\pi}{3}\right)+\mathbf{i}\sin$ $\left(\frac{2k\pi}{3}\right),~k=0,1,2$
how did you come out for those cube roots?it looks very good though...so with it then i find 3 solutions right?

7. 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

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

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

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

Tonio