# Complex Roots

1. Jul 26, 2011

### BloodyFrozen

Does anyone know where I can review finding complex roots?

For example, x3= 8

I know the roots are 2, -1+isqrt(3), and -1-isqrt(3), but I can't remember how to figure it out.

2. Jul 26, 2011

### micromass

Hi BloodyFrozen!

Let me give you the method. The trick is to apply De Moivre's formula. So, the first this to do is to write the complex numbers in polar coordinates. That is

$$z^3=8$$

becomes

$$(r(\cos(\theta)+i\sin(theta)))^3=8(\cos(0)+\sin(0))$$

By De Moivre, we get

$$r^3(\cos(3\theta)+i\sin(3\theta))=8(\cos(0)+\sin(0))$$

So we get that

$$r^3=8,~\cos(3\theta)=\cos(0),~\sin(3\theta)=\sin(0)$$

So

$$r=2, \theta=2\pi k/3~\text{for k an integer}$$

Hence,

$$z=2(\cos(2\pi k/3)+i\sin(2\pi k/3).$$

Thus z=2 if k=0, $z=-1+i\sqrt{3}$ for k=1 and $z=-1-\sqrt{3}$ for k=2. Any other value of k will yield the same values.

3. Jul 26, 2011

### SteveL27

X^3 - 8 = 0

By inspection, x = 2 is a root.

(x^3 - 8) / (x-2) = x^2 + 2x + 4, a quadratic that gives the other two roots.

4. Jul 26, 2011

### BloodyFrozen

Thanks both:)
micromass:That's what I was looking for
Steve:Ahh, right... can't forget about poly division:P

5. Jul 27, 2011

### pmsrw3

$$\begin{eqnarray*} 8 & = & 2^3 e^0 &|& 2^3 e^{2\pi i} &|& 2^3 e^{4\pi i} \\ 8^{\frac{1}{3}} & = & 2 e^0 &|& 2 e^{\frac{2\pi i}{3}} &|& 2 e^{\frac{4\pi i}{3}} \\ & = & 2 &|& 2\left(\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right) &|& 2\left(\cos\left(\frac{4\pi}{3}\right) + i\sin\left(\frac{4\pi}{3}\right)\right) \\ & = & 2 &|& 2\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) &|& 2\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) \\ & = & 2 &|& -1 + i\sqrt{3} &|& -1 - i\sqrt{3} \\ \end{eqnarray*}$$
This is basically what micromass did, but I find it easier to remember by explicitly writing it as a complex exponential. In fact, for many purposes you can stop at step 2 -- $2 e^{\frac{2\pi i}{3}}$ is likely to be as useful or more than $-1 + i\sqrt{3}$.