Roots of unity: solving z^5 = 1

[wduk]

Hello

In my book i have the following with how to solve z^5 = 1

z= r(cos(theta) + i sin(theta)) and 1 = cos(0) + i sin(0)

So if I use De moivre's formula i have:

r^5(cos(5*theta)+ i sin(5*theta) = cos 0 + i sin 0

Now i am not 100% sure if the RHS above was multiplied by 5 as well or remains the same. Since either way we get 0 here....... But say you had:

z^x = -1

So -1 = cos(pi) + i sin(pi)

With de moivres formula does it become:

r^x(cos(x*theta) + i sin(x*theta) = cos(pi*x) + i sin(pi*x)

Or is the RHS unchanged: = cos(pi) + i sin(pi) ?

 

[HallsofIvy]

If find this extremely distressing! You say "In my book i have the following with how to solve z^5 = 1
z= r(cos(theta) + i sin(theta)) and 1 = cos(0) + i sin(0)
So if I use De moivre's formula i have:
r^5(cos(5*theta)+ i sin(5*theta) = cos 0 + i sin 0".

Okay, what is "De Moivre's formula". Is it a formula for a power of a number or does it apply to an entire equation? Frankly, it looks like you are making the mistake of memorizing formulas without learning what they mean!

 

[wduk]

A lot of the subject i have read i have found pretty confusing it rushes through a lot of stuff.

 

[mmm4444bot]

A lot of the subject i have read i have found pretty confusing...

Please post the first statement in your book that you found confusing; we can help explain it. :cool: