Question

Solve the equation. (List your answers counterclockwise about the origin starting at the positive real axis. Express θ in radians.)

z8 + i = 0

Answer 1

\[z^{8}+i=0\]
\[z^{8}=-i\]
Use the fact that any complex number z=Abs[z] * e^(i*phi+i2*pi*k)
\[(\left| z \right| e^{i \phi})^{8}=\left| -i \right| e^{- i \frac{\pi}{2}+i 2\pi k}\]
\[\left| z \right|^{8} e^{8i\phi}=1 e^{- i \frac{\pi}{2} +i2\pi k}\]
So abs[z]^8 =1 so abs[z]=1 and
\[8i\phi=-i \frac{\pi}{2} +i2\pi k\]
\[\phi=- \frac{\pi}{2*8} + \frac{2 \pi k}{8}\]
\[\phi=- \frac{\pi}{16} + \frac{\pi}{4}k\]
Now just enter different values for k (from 0 to 7 since there are 8 solutions) and you get:
k=0, Phi= - Pi/16
k=1, Phi= 3 Pi/16
k=2, Phi= 7 Pi/16
k=3, Phi= 11 Pi/16
k=4, Phi= 15 Pi/16
k=5, Phi= 19 Pi/16
k=6, Phi= 23 Pi/16
k=7, Phi= 27 Pi/16
For your answer (start at real axis) list phi starting with 3 Pi/16

Answer 2

I have a question. How did you get the pi/2? I thought it was pi+2kpi

Answer 3

Oh. The argument ( the angle ) of -i is -pi/2 since the number -i is located directly one unit down form the center of the complex plane.

You could also take +3pi/2 as the angle form -i since that represents same location.