# Thread: Complex number trig conversion.

1. ## Complex number trig conversion.

(1-root 3 i)^6

I can do this but the confusion is the trig part here.
we know that tan theta = b/a

thus tan theta = - root 3
thus theta = - pi/3

But in the solution they subtract this from 2pi to get 5pi/3 as theta.
Why?

2. In what quadrants is $\displaystyle \tan \theta$ negative?

Is there a difference between $\displaystyle \displaystyle \frac{-\pi}{3}$ and $\displaystyle \displaystyle \frac{5\pi}{3}$ ?

3. Originally Posted by pickslides
In what quadrants is $\displaystyle \tan \theta$ negative?

Is there a difference between $\displaystyle \displaystyle \frac{-\pi}{3}$ and $\displaystyle \displaystyle \frac{5\pi}{3}$ ?
Right, they are the same thing.
okay so one last thing. I got the answer as:

2^6 = r
so its r(cos n theta + i sin n theta)
I get:

64(cos 2pi + i sin 2pi) or if you use 5pi/3 its 64(cos 10pi + i sin 10pi)

but sin 2pi is 0, so the answer should just be 64. The solution shows its 64+64i, is that a mistake?

4. The correct answer is 64, so you are correct.