# Homework Help: Z^3-1=0 (express using de moivres theorum)

1. Oct 17, 2012

### Daaniyaal

1. The problem statement, all variables and given/known data

Use De Moivre's theorem to obtain solutions to the equation (Z^3)-1=0

2. Relevant equations

z^n= rθ^n(cosnθ+isinnθ)

3. The attempt at a solution

I am having difficulty on figuring out how to meld the equation to De Moivres Theorem

2. Oct 17, 2012

### Dick

Write z=r(cos(θ)+isin(θ)). So you have z^3=1. What are the possibilities for r and θ? Start with r. By convention we pick r>=0. What must r be?

Last edited: Oct 17, 2012
3. Oct 17, 2012

### Daaniyaal

r would be 1 and θ=0 right?

4. Oct 17, 2012

### Dick

Good start! That's one solution. Can you find two more? Try to think of other values of θ that might work. cos(3θ) has to be 1 and sin(3θ) has to be 0.

Last edited: Oct 17, 2012
5. Oct 17, 2012

### Daaniyaal

ok so

z^3= 1^3(cos(3θ)+isin(3θ))

6. Oct 17, 2012

### Dick

Not so special. cos(2π)=1 and sin(2π)=0. Suggest a value for θ using that.

7. Oct 17, 2012

### Daaniyaal

2pi!

8. Oct 17, 2012

### Daaniyaal

Sorry I keep taking too many hints, I get it after I post that I don't understand

9. Oct 17, 2012

### HallsofIvy

The number "1" has "polar form" r= 1 and $\theta= 0$ because the number 1= 1+ 0i is represented by the point (1, 0) which is at distance 1 from (0, 0) and angle 0 with the positive x-axis. Alternatively, x= 1+ 0i= 1(cos(0)+ i sin(0)).

Now, cosine and sign are periodic with period $2\pi$. So the x-axis, in addition to being "$\theta= 0$" is also "$\theta= 2\pi$" and "$\theta= 4\pi$".

(As well as "$6\pi$" or "$9\pi$" or even "$-2\pi$", etc. but those are not important. Do you see why?)

10. Oct 17, 2012

### Daaniyaal

Because it is really the same position right? 2pi=4pi in terms of position on the circle?

11. Oct 18, 2012

### Dick

True. But we are mainly interested in values of θ between 0 and 2π. The others are really just 'copies'. If you put θ=2π you get the same z value as θ=0. I was hoping you'd say 2π/3. Do you see why that works and is more interesting?