Faster way to for $z^3 = -2 (1+i \sqrt 3) \bar z$ than complex algebra

Question

What is the fastest way to solve for $z^3 = -2 (1+i \sqrt 3) \bar z$?

I know how to do this using complex algebra. but that takes a long time. Can someone show me a faster way?

Answer 1

Multiply by $z$ to obtain $z^4 = -2 (1+i\sqrt 3)|z|^2$. Looking at absolute values we see that $|z|^2 = \sqrt{4 + 4 \cdot 3} = \sqrt{16} = 4$, finally looking at angles we have $$4 \arg z = \arg (-1-i\sqrt 3) = -\frac{2\pi}3$$ Combined we get $z = |z| e^{i\arg z} = 2 e^{-i\frac{\pi}6 + k \frac\pi2} = (1-\sqrt 3i) \cdot e^{\frac{ik\pi}2}$ where $k\in\{0,1,2,3\}$ so all in all $$z = \pm(1\pm \sqrt3 i)$$ And of course the trivial solution $z=0$.

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Note that I used a few equalities: $\arg zw \equiv \arg z + \arg w \pmod{2\pi}$, $|z| = \sqrt{\Re^2 z + \Im^2 z} = \sqrt{z\bar z}$ and $\exp(i\frac\pi6) = \frac12(1+\sqrt3 i).$

Answer 2

Taking the modulus, $|z|^3=2\cdot2\cdot|z|$, and $|z|=0$ or $|z|=2$ ($|z|=-2$ cannot hold).

Taking the argument, $3\arg z=\pi+\pi/3-\arg z+2k\pi$, and $\arg z=\pi/3+k\pi/2$.

$$z=0,1+i\sqrt3,-\sqrt3+i,-1-i\sqrt3,\sqrt3-i.$$