Question

Use trigonometric form and De Moivre's thereom to simplify the expression. Put in form a+bi rounding to one decimal place.
(-3+3i(square root of 3))^2

Answer 1

\[(-3+3i \sqrt{3})^2\]

Answer 2

Well,
\[-3 + 3i\sqrt{3} = 6\left( \frac{-1}{2} + i\frac{\sqrt{3}}{2}\right) = 6\left(\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3}) \right) \]
What does De Moivre's theorem say?

Answer 3

\[z=r(\cos \theta+isin \theta)
z^n=[r(\cos \theta+isin \theta
=r^n(\cos n \theta+isin n \theta)\]

Answer 4

okay, so apply that to the above expression. What do you get?

Answer 5

I don't even understand honestly... I didn't even know where to start. I've been on bed rest and I was given problems like these and I have no idea. I just need explanation on how to do it, then it will come easy.

Answer 6

Mm, I'm sorry to hear that, I hope you feel better soon.

As you said, De Moivre's theorem says
\[ \left(\cos(\theta) + i\sin(\theta)\right)^n = \cos(n\theta) + i\sin(n\theta) \]
in the above problem theta = 2 pi / 3 and n = 2. Therefore, that expression equals
\[ 36\left( \cos(\frac{4\pi}{3}) + i\sin(\frac{4\pi}{3})\right) = 36\left(\frac{-1}{2} + -i\frac{\sqrt{3}}{2}\right) \]
\[ = -18- 18i\sqrt{3} \]

Answer 7

where'd n=2 come from?

Answer 8

and thank you, I hope I do too.. I've been sick for over a month now and it is no fun.

Answer 9

Also, the program says it is a wrong answer. :/

Answer 10

The 2 comes from the fact that you're squaring the term in parentheses. And having double checked, I am quite confident that that's the answer to the problem you wrote down, are you sure you didn't make any typos? :)

Backup: http://www.wolframalpha.com/input/?i=%28-3+%2B+3i*sqrt%283%29%29^2

Answer 11

I'm positive it's typed correctly.. I figured it out you were correct however the imaginary numbered needed to be in decimal form. so
-18[\sqrt{3}= -18-31.2i\]

Answer 12

Ah, I see. Well I'm glad it worked out!

Answer 13

Me too, thanks again:)