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)

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 $\displaystyle (1-\sqrt{3}\,\mathbf{i})^4$

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

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

Now: $\displaystyle \left[ {2\left( {\cos \left( {\frac{{ - \pi }}{3}} \right) + i\sin \left( {\frac{{ - \pi }}{3}} \right)} \right)} \right]^4 = 16\left( {\cos \left( {\frac{{ - 4\pi}}{3}} \right) + i\sin \left( {\frac{{ - 4\pi }}{3}} \right)} \right)$

Re: A couple of beginner complex number questions

Ah brilliant thanks a bunch! i have been trying to get around this for ages!

(Rofl)