# complex number equation

• Sep 24th 2011, 11:24 AM
elieh
complex number equation
I'm trying to get the 4 solutions for this:
$w^4 = -8 +(8*3^\frac{1}{2})i$

In cis form and after applying de Moivre's theorem I get:

$2cis( \frac{-\pi}{12} +\frac{1}{2}k\pi)$

Where k is varying integer.

I've checked several times and I can't see where I'm wrong.
Help appreciated!
• Sep 24th 2011, 11:42 AM
Plato
Re: complex number equation
Quote:

Originally Posted by elieh
I'm trying to get the 4 solutions for this:
$w^4 = -8 +8*3^\frac{1}{2}$

In cis form and after applying de Moivre's theorem I get:

$2cis( \frac{-\pi}{12} +\frac{1}{2}k\pi)$

Where k is varying integer.

I've checked several times and I can't see where I'm wrong.
Help appreciated!

Have you written that correctly?
OR should it be $w^4 = -8 +8\sqrt3\color{red}\mathbf{i}$
• Sep 24th 2011, 11:56 AM
elieh
Re: complex number equation
I edited it
• Sep 24th 2011, 12:01 PM
Plato
Re: complex number equation
Quote:

Originally Posted by elieh
I'm trying to get the 4 solutions for this:
$w^4 = -8 +(8*3^\frac{1}{2})i$
$2cis( \frac{-\pi}{12} +\frac{1}{2}k\pi)$

You have the argument wrong.
The only thing to add is $k=0,1,2,3$
• Sep 24th 2011, 12:12 PM
elieh
Re: complex number equation
But I don't get whole numbers to put in my answer which should be in the form a+bi
• Sep 24th 2011, 12:43 PM
Plato
Re: complex number equation
Quote:

Originally Posted by elieh
But I don't get whole numbers to put in my answer which should be in the form a+bi

$\text{Arg}\left( { - 8 + 8\sqrt 3 i} \right) = \pi + \arctan \left( {\frac{{8\sqrt 3 }}{{ - 8}}} \right) = \frac{{2\pi }}{3}$

Now $\frac{2\pi}{3}\frac{1}{4}=\frac{\pi}{6}$
• Sep 24th 2011, 01:01 PM
elieh
Re: complex number equation
That works, Thanks.