Complex roots and de Moivre's formula

[Jonmundsson]

Homework Statement

Find all the roots of z^{6} = -2

Homework Equations

z = \sqrt[n]{r} (cos(\frac{\theta}{n} + k\cdot \frac{2\pi}{n}) + isin(\frac{\theta}{n} + k\cdot \frac{2\pi}{n}) where k = 0, 1, \cdots, n - 1

The Attempt at a Solution

I know that r = \sqrt{x^2 + y^2} = \sqrt{(-2)^2} = 2 but I don't understand how I find \theta. arg z = arg -2 isn't a nice number to work with and now I'm stumped. This is the only thing holding me back since the formula is pretty straightforward so if anyone could tell me how I find \theta that would be great.

Thanks in advance.

 

[Hootenanny]

Homework Statement

Find all the roots of z^{6} = -2

Homework Equations

z = \sqrt[n]{r} (cos(\frac{\theta}{n} + k\cdot \frac{2\pi}{n}) + isin(\frac{\theta}{n} + k\cdot \frac{2\pi}{n}) where k = 0, 1, \cdots, n - 1

The Attempt at a Solution

I know that r = \sqrt{x^2 + y^2} = \sqrt{(-2)^2} = 2 but I don't understand how I find \theta. arg z = arg -2 isn't a nice number to work with and now I'm stumped. This is the only thing holding me back since the formula is pretty straightforward so if anyone could tell me how I find \theta that would be great.

Thanks in advance.

Welcome to Physics Forums.

Perhaps it would be easier if you wrote the equation in exponential notation:

r^6 e^{i6\theta} = 2 e^{i\pi}

Can you take it from here?

 

[Jonmundsson]

Wow! Of course! Thanks for the quick help!

 

[Hootenanny]

Wow! Of course! Thanks for the quick help!

A pleasure