# Thread: Roots of an equation

1. ## Roots of an equation

$\varphi'''-\lambda\varphi=0$

$m^3-\lambda=0\Rightarrow m^3=\lambda$

My book then does this:

$\displaystyle m=s, \ m=\left(\frac{-1\pm i\sqrt{3}}{2}\right)s,\ \ s=\lambda^{1/3}$

How were the roots of m determined?

2. $\displaystyle w_k=r^{1/n}\left[\cos\left(\frac{0+2\pi k}{n}\right)+i\sin\left(\frac{0+2\pi k}{n}\right)\right]$

$r=\lambda^{1/3}=s$

Solved.

3. Right. Just the third roots of unity.