# Thread: Complex Roots of Unity

1. ## Complex Roots of Unity

Hello, I am having trouble drawing the unit circle in complex roots of unity.

For exaple the question: z^4 = -16

I have worked out that r = 2, and that each angle is pi/2. How do i Solve the question? I thought that z1 would equal -1, but it does not. Can someone please explain how to do it?
EDIT: it was actually z^4 = -16

2. Originally Posted by noobonastick
Hello, I am having trouble drawing the unit circle in complex roots of unity.

For exaple the question: z^4 = 16

I have worked out that r = 2, and that each angle is pi/2. How do i Solve the question? I thought that z1 would equal -1, but it does not. Can someone please explain how to do it?
Hi

Let $z = r\:e^{i\theta}$ then $z^4 = r^4\:e^{4i\theta}$

$r^4\:e^{4i\theta} = 16$ implies
$r^4 = 16$ and $4\:\theta = 2k\pi$

which means $r = 2$ and $\theta = k\frac{\pi}{2}$

k=0 => z=2
k=1 => z=2i
k=2 => z=-2
k=3 => z=-2i

3. Sorry we use the form cos(theta) +isin(theta). Could you explain it in those terms.
(i also edited my original post)
EDIT: dont worry i have figured it out. Thanks

4. Originally Posted by noobonastick
Sorry we use the form cos(theta) +isin(theta). Could you explain it in those terms.
(i also edited my original post)
EDIT: dont worry i have figured it out. Thanks
What are you not seeing?

You have $\theta = k \cdot \frac{\pi}{2}$.

Substitute k = 0, 1, 2, and 3

It matters not what notation you use.

5. -16 is at $\theta= \pi$ as measured from the positive x-axis. The first root (the principle fourth root of unity) will be at $\pi/4$ and the other three will be $\pi/4+ \pi/2= 3\pi/4$, $3\pi/4+ \pi/2= 5\pi/4$, and [tex]5\pi/4+ \pi/2= 7\pi/4[/itex] on the circle of radius 2.