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$w$ is a complex number that satisfy: $$5{w^3} - 3i{\left| w \right|^2} - 2i=0$$ Can we find $|w|$ without finding $w$ or easy root?

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Take $w = r e^{i \phi}$. Then you get:

$$5 r^3 e^{i 3 \phi} - 3i r^2 - 2i = 0$$

Which can be rearranged to:

$$5r^3 e^{i 3 \phi} = i(3r^2 + 2)$$

divide by $i$ on both sides:

$$5r^3 e^{i 3 \phi - i\frac{\pi}{2}} = \underbrace{3r^2 + 2}_{\in \mathbb R}$$

since RHS is real and positive, LHS is real and positive, i.e. $e^{ i 3 \phi - i \frac{\pi}{2}} = 1$. Thus we get:

$$5r^3 - 3r^2 - 2 = 0$$

and we want the positive real roots to this equations, which is $r = 1$.

EDIT: Fixed an error.

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The answer is yes we can...

For any solution $w$, the two complex numbers $5w^3$ and $(3\left| w \right|^2+2)\mathrm{i}$ are the same hence their modulus are the same: $5\left|w\right|^3=3\left| w \right|^2+2$. This means that the modulus $r$ of any solution $w$ is such that $5r^3-3r^2-2=0$. Here ends the answer to the OP question.

Naturally one can go further since an obvious solution of the $r$ equation is $r=1$. Hence $5r^3-3r^2-2=(r-1)(5r^2+2r+2)$ and, since $5r^2+2r+2=0$ has no real solution, every solution of the $w$ equation is such that $\left| w \right|=1$.

All this does not say that there exists any solution $w$ at all but one can go further and note that a complex number $w$ such that $\left| w \right|=1$ is a solution iff the argument of $w^3$ is $\pi/2$ hence the argument of $w$ is $\pi/6$ or $5\pi/6$ or $3\pi/2$.

Finally the three solutions are $\mathrm{e}^{\mathrm{i}\pi/6}=(\sqrt{3}+\mathrm{i})/2$, $\mathrm{e}^{\mathrm{i}5\pi/6}=(-\sqrt{3}+\mathrm{i})/2$ and $\mathrm{e}^{\mathrm{i}3\pi/2}=-\mathrm{i}$.

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Since the second and third terms are pure imaginary, the angle of the solution is either $\pi/6$, $5\pi/6$, or $3\pi/2$. To solve for the magnitudes, just solve the real equation

$$5{w^3} - 3 w^2 - 2=0$$

which clearly has a real positive root at $1$.

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