# A couple of beginner complex number questions

• Oct 2nd 2011, 05:48 AM
ShippingY
A couple of beginner complex number questions
Hello guys, im new here, this seems like a nice forum to help me catch up on my math (Nod) I havent been doing math for years and now I started uni and I got a lot to catch up.

I cant seem to manage to get the math format in so I apologize in advance for the confusing syntex

So I got a couple of questions I did not manage to get past in complex numbers.
If you could explain how to break it down to a newbie i would be in your debt,
give the trigonometric form [(r=cos(theta) + isin(theta)] of:
(1-sqrt(3i))^4
when i break it down the long way i get to
=-8-21*sqrt(2)+18i-4*sqrt(3i)
and from there its a prety dead end, r= 222*sqrt(2) which i believe is probably incorrect

the other problem is a little simpler but I got stuck with the isin value
sqrt(3)+3i
now i broke it down to r = 2*sqrt(3)
and from there cos(theta)=sqrt(3)/2*sqrt(3) * sqrt(3)/sqrt(3) = 1/2
but for the isin i cant get past the isin=3/2sqrt(3)
how do i get rid of this pesky square root 3? i know that it is either positive or negative sqrt(3)/2 according to the unit circle, but how do i find which?

Thanks a bunch
Yshipping (Bow)
• Oct 2nd 2011, 10:23 AM
Plato
Re: A couple of beginner complex number questions
Quote:

Originally Posted by ShippingY
give the trigonometric form [(r=cos(theta) + isin(theta)] of:
(1-sqrt(3i))^4

Surely it is $(1-\sqrt{3}\,\mathbf{i})^4$
First step: $|1-\sqrt{3}\,\mathbf{i}|=\sqrt{1^2+(\sqrt{3})^2}=2$.

Second step: $\text{Arg}(1-\sqrt{3}\,\mathbf{i})=\frac{-\pi}{3}$.

Now: $\left[ {2\left( {\cos \left( {\frac{{ - \pi }}{3}} \right) + i\sin \left( {\frac{{ - \pi }}{3}} \right)} \right)} \right]^4 = 16\left( {\cos \left( {\frac{{ - 4\pi}}{3}} \right) + i\sin \left( {\frac{{ - 4\pi }}{3}} \right)} \right)$
• Oct 2nd 2011, 11:03 AM
ShippingY
Re: A couple of beginner complex number questions
Ah brilliant thanks a bunch! i have been trying to get around this for ages!
(Rofl)