# Difficult complex number question

• August 13th 2010, 04:44 AM
Glitch
Difficult complex number question
The question:
Let $z = (1 + \sqrt{3} i)$ and w = (1 + i). Find Arg(z) and Arg(w) and hence Arg(zw). Evaluate zw and hence show that $cos\frac{7\pi}{12} = \frac{1 - \sqrt{3}}{2\sqrt{2}}$. Find a similar expression for $sin \frac{7\pi}{12}$

I found:
Arg(z) = $\frac{\pi}{3}$
Arg(w) = $\frac{\pi}{4}$
Arg(zw) = $\frac{\pi}{12}$
zw = $2\sqrt{2}e^{i\frac{\pi}{12}}$

The last bit of the question is giving me trouble. I tried using de Moivre's theorem as follows:

$(2\sqrt{2}e^{i\frac{\pi}{12}})^7$
$(2\sqrt{2})^7e^{i\frac{7\pi}{12}}$

This gives me the argument I want. But I'm unsure of how to solve this. Changing it to sin/cos form (not sure of the proper name for it) seems to be the way to go, but I'm still stumped:

$(2\sqrt{2})^7(cos(\frac{7\pi}{12}) + isin(\frac{7\pi}{12}))$

Any assistance would be great!
• August 13th 2010, 04:55 AM
Ackbeet
I agree with arg(z) and arg(w). However, I don't agree with your arg(zw). How do arguments behave under multiplication?
• August 13th 2010, 04:59 AM
Glitch
Oh whoops, that should be 7Pi/12.

I guess that makes zw:

$2\sqrt{2}e^{i\frac{7\pi}{12}}$

I'm still not sure how to solve it though.
• August 13th 2010, 05:01 AM
Ackbeet
Try multiplying zw in both the polar form and the rectangular form, and then setting them equal. What does that give you?
• August 13th 2010, 05:13 AM
Glitch
Ahh! Got it! Thank you very much. :)

This is what I did, for anyone that's looking on:

zw = $(1-\sqrt{3}) + i(1 + \sqrt{3})$ (rectangular form)
zw = $2\sqrt{2}(cos(\frac{7\pi}{12}) + isin(\frac{7\pi}{12}))$

Equating real parts:
$2\sqrt{2}(cos(\frac{7\pi}{12})) = 1-\sqrt{3}$
Therefore,
$cos\frac{7\pi}{12} = \frac{1 - \sqrt{3}}{2\sqrt{2}}$
• August 13th 2010, 05:14 AM
Ackbeet
Looks good to me. You're very welcome.