# Precal hw on complex numbers

1. Nov 24, 2004

### MercuryRising

Z^6 + 64 = 0
there are suppose to be 6 solutions
need help on how to find the roots

Last edited: Nov 24, 2004
2. Nov 24, 2004

### vsage

Are you familiar with de Moivre's theorem? The problem is pretty easy then. Alternatively, you can factor the equation $$Z^6 + 64=0$$ into
$$(Z^3 - 8i)(Z^3+8i)$$ from which you can factor more.

Last edited by a moderator: Nov 24, 2004
3. Nov 24, 2004

### MercuryRising

opps, sorry, i must 've made myself unclear, the goal is to find the 6 values of z, but thanks for heling on the roots!

4. Nov 24, 2004

### Gokul43201

Staff Emeritus
Write $z^6 = r^6(cos 6\theta + isin 6\theta)$ and solve.

5. Nov 25, 2004

### nolachrymose

Vsage's point was that, from there, you can factor further, thus getting the six linear roots (hint: think about sum and difference of cubes).