Question

Find the fourth roots of 256(cos 240° + i sin 240°)

Answer 1

like a first step 256 = 4*4*4*4

Answer 2

let z=256(cos 240+i sin 240)
\[=4^4(\cos 240+i \sin 240)=4^4e ^{\iota 240}=4^4e ^{\iota(240+360n)}\]
\[z ^{\frac{ 1 }{ 4 }}=4 e ^{\iota(\frac{ 240+360 n }{ 4 })}=4e ^{\iota (60+90n)}\]
put n=0,1,2,3
you get four fourth roots of z

Answer 3

4th root of 256 is 4, and 240/4 = 60, and 360/4 = 90, so the 4 roots are
4(cos 60 + i sin 60)
4(cos 150 + i sin 150)
4(cos 240 + i sin 240)
4(cos 330 + i sin 330

Answer 4

correct.

Answer 5

does that help?

Answer 6

What are "e" and "l" in the formula?

Answer 7

\[\large\rm e^{i \theta}=\cos \theta+i \sin \theta\]If you haven't learned about the complex exponential yet, that's fine.
We can stay in sines and cosines so it makes more sense to you.

Answer 8

The important thing to note though,
is that you must adjust your angle to allow for rotations BEFORE apply De'Moivre's Theorem.
\[\large\rm \cos(240)=\cos(240+360)=\cos(240+2\cdot360)\]If we spin a full 360 around the circle, we can land at that same spot as before, ya?
So adding 360 doesn't change anything.
These are all coterminal.
Another way we can write this,\[\large\rm \cos(240)=\cos(240+360k)\]Where k is any positive or negative whole number.
Any full number of spins can be added or subtracted and give us the same result.

Answer 9

So that's why we have to rewrite our expression as,\[\rm 256\left[\cos(240+i \sin(240)\right]\quad=\quad 256\left[\cos(240+360k)+i \sin(240+360k)\right]\]before we apply De'Moivre's Theorem.

Answer 10

The rest of the problem is explained above by these guys.
Hopefully that fills in any details that you're missing.