Complex roots of a cubic equation

[engineer1406]

Homework Statement

Hi all,

I was wondering if there is a procedure you can follow to calculate the complex roots of a cubic equation.

Homework Equations

For example the equation

x³ - 1 = 0

has roots of x = 1
x = -0.5 + √3/2 i
x = -0.5 - √3/2 i

Admittedly, I got those solutions off wolfram alpha, but I am wondering how to work it out without wolfram!

Thanks!

The Attempt at a Solution

 

[micromass]

In this case, it's not so hard. If you have a complex number z, then you first need to write it in the form

$$z=R(\cos(\theta)+i\sin(\theta))$$

It is now a theorem (prove this!), that the n-th roots are exactly

$$\sqrt[n]{R}(\cos(\frac{\theta+2k\pi}{n})+ i\sin(\frac{\theta+2k\pi}{n}))^n$$

for $0\leq k<n$. This is due to De Moivre's identity.

So, can you use this information to calculate the third roots of 1?

 

[engineer1406]

Haha, great. The formula works!

Thanks very much micromass!

 

[HallsofIvy]

Here's another way to do it (not as "sophisticated"): $x^3- 1= 0$ has the obvious solution x= 1 so x- 1 is a factor. Dividing $x^3- 1$ by x- 1, we find that $x^3- 1= (x- 1)(x^2+ x+ 1)$. If x is not 0 then $x^2+ x+ 1= 0$. That's a quadratic equation so use the quadratic formula to solve it.