# Math Help - Difficult complex number question

1. ## 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!

2. I agree with arg(z) and arg(w). However, I don't agree with your arg(zw). How do arguments behave under multiplication?

3. 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.

4. Try multiplying zw in both the polar form and the rectangular form, and then setting them equal. What does that give you?

5. 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,
$cos\frac{7\pi}{12} = \frac{1 - \sqrt{3}}{2\sqrt{2}}$

6. Looks good to me. You're very welcome.