Short and sweet, the problem to solve is:
$$(\overline{z}-i)^4 = \cos{\frac{4\pi}{3}} +i\sin{\frac{4\pi}{3}}$$
My process of solving was as follows:
- First, take the fourth root using the polar coordinate representation of complex numbers as follows: $$\overline{z}-i = r^\frac{1}{4}e^{\frac{\phi+2k\pi}{4} i} $$
Note that there are four solutions for $k=0,1,2,3$, but for the sake of concision, I will only do the case of $k=0$. The rest can be calculated with similar methods.
$$\overline{z}-i = \frac{1}{2} +i\frac{\sqrt{3}}{2}$$
- Then, move $i$ to the right side and get
$$\overline{z} = \frac{1}{2} + \large i(\frac{\sqrt{3}}{2}+1) $$
- Lastly, since we have $z$ conjugated on the left side, take the conjugate of the right side and get:
$$z = \frac{1}{2} - \large i(\frac{\sqrt{3}}{2}+1) $$
Can someone please verify if this is correct? If not, what is the correct method to do this?
Once again, I am aware that this equation has four possible solutions in the complex plane.