question asks:
(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?
question asks:
(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?
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?