Find the 8th complex roots of $-81i$.

Question

Find the 8th complex roots of $-81i$.

So $i = \cos(\pi/2) + i\sin(\pi/2) = \cos(5\pi/2) + i\sin(5\pi/2)$

thus $z^8 = -81e^{(\pi + 4k\pi)/2 * i}$ So,

$z = \sqrt{3} \cdot e^{i \cdot \frac{3\pi + 4k\pi}{16}}$

Confirm?

Answer 1

That looks correct.

If you want to check this, you can also raise it to the eight power yourself to obtain $$z^8 = \left(\sqrt{3} \cdot e^{i \cdot \frac{3\pi + 4k\pi}{16}}\right)^8 = 81 \cdot e^{8\left(i \cdot \frac{3\pi + 4k\pi}{16}\right)} = 81 \cdot e^{i \cdot \frac{3\pi + 4k\pi}{2}} = 81 \cdot e^{i \cdot \frac{3\pi}{2}} = 81 \cdot (-i) = -81i$$