How to solve 4th degree polynomial equation with complex coefficients numerically?

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I have a polynomial equation $-(a-ib)e^{(4\pi i/3)}(\sqrt{2}i+x^3/\sqrt{3})x- (a+ib) e^{(2\pi i/3)}(\sqrt{2}ix^3+1/\sqrt{3})=0$ with the conditions

$a^{2}+b^{2} \leq 1$,

$1/2 \geq a \geq -1$, and

$\sqrt{3}/2 \geq b \geq -\sqrt{3}/2$.

I want numerical solutions for this equation. I tried in Mathematica, but ended up with messy solutions. Also, I couldn't impose conditions on $a$ and $b$. Please help me in solving this.

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It is not clear if you are seeking over $x \in \mathbb{R}$ or $x \in \mathbb{C}$. Nevertheless here are some particular cases.

  • Suppose $a=b=0$. Then any $x \in \mathbb{C}$ is a solution.

  • Suppose $(a+ib) = (\rho i)e^{-2 \pi i/3}$ for some $\rho \in \mathbb{R}$, $\rho \neq 0$. Then $$ (a+ib)e^{2\pi i /3} = \rho i \quad , \quad (a-ib)e^{4 \pi i /3} = -\rho i $$ So your equation reduces to $$ (ix\sqrt{2} + x^4/\sqrt{3}) = (ix^3\sqrt{2} + 1/\sqrt{3})$$ which is satisfied when $x=1$.

  • Suppose $(a+ib) = \rho e^{-2\pi i/3}$ for some $\rho \in \mathbb{R}$, $\rho \neq 0$. Then $$ (a+ib)e^{2\pi i /3} = \rho \quad , \quad (a-ib)e^{4 \pi i /3} = \rho $$ So your equation reduces to $$ (ix\sqrt{2} + x^4/\sqrt{3}) = -(ix^3\sqrt{2} + 1/\sqrt{3})$$ which has no solution over $x \in \mathbb{R}$ because it would require $x^4 = -1$.