Z^3-1=0 (express using de moivres theorum)

[Daaniyaal]

Homework Statement

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

Homework Equations

z^n= rθ^n(cosnθ+isinnθ)

The Attempt at a Solution

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

 

[Dick]

Homework Statement

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

Homework Equations

z^n= rθ^n(cosnθ+isinnθ)

The Attempt at a Solution

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

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?

 

[Daaniyaal]

Write z=r(cos(θ)+isin(θ)). So you have z^3=1. What are the possibilities for r and θ?

r would be 1 and θ=0 right?

 

[Dick]

r would be 1 and θ=0 right?

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.

 

[Daaniyaal]

Good start! That's one solution. Can you find two more, try to think of other values of θ that might work.

ok so

z^3= 1^3(cos(3θ)+isin(3θ))

 

[Dick]

Arghh I can't think of anything, do they have anything to do with our "special triangles"? the 45-45-90 one and the 30-60-90 one? I've thought of multiple sin and cos values and none of them are giving me the right answer :( apart from the previous one I have listed.

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

 

[Daaniyaal]

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

2pi!

 

[Daaniyaal]

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

 

[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?)

 

[Daaniyaal]

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?)

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

 

[Dick]

2pi!

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?