Reviewing How to Find Complex Roots: e.g. x3= 8

[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.

 

[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.

 

[SteveL27]

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.

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.

 

[BloodyFrozen]

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

 

[pmsrw3]

$$\begin{eqnarray*}<br /> 8 & = & 2^3 e^0 &|& 2^3 e^{2\pi i} &|& 2^3 e^{4\pi i} \\<br /> 8^{\frac{1}{3}} & = & 2 e^0 &|& 2 e^{\frac{2\pi i}{3}} &|& 2 e^{\frac{4\pi i}{3}} \\<br /> & = & 2 &|& 2\left(\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right) &|& 2\left(\cos\left(\frac{4\pi}{3}\right) + i\sin\left(\frac{4\pi}{3}\right)\right) \\<br /> & = & 2 &|& 2\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) &|& 2\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) \\<br /> & = & 2 &|& -1 + i\sqrt{3} &|& -1 - i\sqrt{3} \\<br /> \end{eqnarray*}$$
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 $-1 + i\sqrt{3}$.