# 2 from 1.2 of basic complex analysis 3rd edition (marsden & hoffman)

just wanted to make sure I'm doing this right

1. solve the following equations

a) $z^6 + 8 =0$

first I write out the complex number -8 in polar form

$-8 = 8(\cos(\pi) + i\sin(\pi))$

applying de moivre's formula I come out with

$z= \sqrt {8} \left(\cos(\pi/6 + 2\pi k/6) + i\sin(\pi/6 + 2\pi k/6)\right)$

• Looks good. Another way could be to use the factorization of $(z^2)^3+2^3$ – voldemort Jan 16 '14 at 0:59

I think it looks good. You might want to use this as well (if you don't remember De Moivre's formula):

• You have:

$$z^6 = -8$$

• Making $z = M e^{i \alpha}$, it yields:

$$M^6 e^{i 6 \alpha} = 8 e^{i \pi}, \quad 0<\alpha < 2 \pi,$$

• Then it follows that:

$$M = 8^{1/6} \quad \wedge \quad 6\alpha = \pi + 2k \pi, \ k = \{0,\ldots, 5 \},$$

• So:

$$z_k = 8^{1/6} e^{i (\pi + 2k \pi)/6 },$$

which are the vertices of an hexagon in the plane $(x,y)$, $x = \text{Re}(z), y = \text{Im}(z)$, and matches your result.

I hope this is useful to you.

Cheers!