# Finding Cartesian form from argument and modulus

• July 31st 2010, 02:58 AM
Glitch
Finding Cartesian form from argument and modulus
The question:
Find the "a + ib" form of the complex numbers whose modulus and principle argument are:

|z| = 3, arg(z) = Pi/8

My attempt:

$3(cos(\frac{\pi}{8}) + isin(\frac{\pi}{8}))$

I'm certain they want the solution as an exact value, so Pi/8 is going to be a pain to find. I'm thinking I have to use the double-angle formulas, but I'm doing something wrong...

$cos(2x) = cos^2x - sin^2x$
$cos (\frac{\pi}{8}) = cos^2(\frac{\pi}{16}) - sin^2(\frac{\pi}{16})$

Clearly that isn't any easier. Have any suggestions? Thanks.
• July 31st 2010, 03:39 AM
Prove It
You'll actually need the half-angle formulas

$\sin{\frac{\theta}{2}} = \pm \sqrt{\frac{1 - \cos{\theta}}{2}}$

and

$\cos{\frac{\theta}{2}} = \pm \sqrt{\frac{1 + \cos{\theta}}{2}}$.

In this case, $\theta = \frac{\pi}{4}$.
• July 31st 2010, 03:43 AM
Glitch
Heh, I didn't realise half-angle formulas existed! Thanks!