# Roots of an equation

• Aug 12th 2009, 10:45 AM
arbolis
Roots of an equation
I must find the root of $z^3+1=0$.
I both used the polar and Cartesian approaches and reached the result that the root is $z=-1$.
Is it possible? Shouldn't there be 3 roots? (they could be multiple roots I guess).
I can show the details of my attempt if you like.
• Aug 12th 2009, 10:50 AM
running-gag
You have found only one root !
Now you can factor $z^3+1 = (z+1)(z^2-z+1)$ and solve $z^2-z+1=0$ to find the two other roots
• Aug 12th 2009, 10:56 AM
running-gag
Anyway you should have found 3 roots

$z^3 = -1$

$r^3 \:e^{3i \theta} = e^{i \pi}$

$r^3=1$ and $3 \theta = \pi + 2k \pi$

$r=1$ and $\theta = \frac{\pi}{3} + \frac{2k \pi}{3}$

$z_1 = e^{i \frac{\pi}{3}}$

$z_2 = e^{i \pi} = -1$

$z_3 = e^{i \frac{5 \pi}{3}} = e^{-i \frac{\pi}{3}}$