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:

  1. 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.

$$\overline{z}-i = \frac{1}{2} +i\frac{\sqrt{3}}{2}$$

  1. Then, move $i$ to the right side and get

$$\overline{z} = \frac{1}{2} + \large i(\frac{\sqrt{3}}{2}+1) $$

  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.

  • $\begingroup$ Yes, it is correct. $\endgroup$ – Yves Daoust Dec 4 '20 at 18:53

$$\bar z-i=e^{(2n\pi i+i4\pi/3)/4}=e^{i\pi(3n+2)/6}$$ where $n=0,1,2,3$

$$\implies \bar z=i+\cos\dfrac{\pi(3n+2)}6+i\sin\dfrac{\pi(3n+2)}6$$

$$\implies z=\bar{\bar z}=-i+\cos\dfrac{\pi(3n+2)}6-i\sin\dfrac{\pi(3n+2)}6$$


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