3
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ Looks good. Another way could be to use the factorization of $(z^2)^3+2^3$ $\endgroup$ – voldemort Jan 16 '14 at 0:59
1
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.