Complex Number Locus w.r.t Coordinate Axes 
Show that the locus formed by z in the equation $z^3 + iz = 1$  never
crosses the co-ordinate axes in the Argand plane. Further show that
$$|z| = \sqrt{\dfrac{-\Im(z)}{2\Re(z)\Im(z) +1}}$$

Concept:
$z=x+iy$
$z^3=(x+iy)^3=x^3-3xy^2+i(3x^2y-y^3)$
$iz=-y+ix$
Combined equation is $x^3-3xy^2-y-1+i(3x^2y-y^3+x)=0$
How do we proceed from here onward?
 A: 
the locus formed by z in the equation $z^3 + iz = 1$  never crosses the co-ordinate axes

Suppose the solution set crossed the real axis $z=x \in \mathbb{R}\,$, then $x^3+ix=1\,$. Then, equating the real and imaginary parts of the two sides gives $x^3=1$ and $x=0\,$, but the system formed by those two equations has no solutions. A similar argument works for the imaginary axis.

$|z| = \sqrt{\frac{-\operatorname{Im}(z)}{2\operatorname{Re}(z)\operatorname{Im}(z) +1}}$

Using that $\operatorname{Re}{z}=(z+\bar z)/2$ and  $\operatorname{Im}{z}=(z-\bar z)/2i\,$:
$$
\begin{align}
|z|^2 = \frac{-\operatorname{Im}(z)}{2\operatorname{Re}(z)\operatorname{Im}(z) +1} \;\;&\iff\;\; z \bar z  = \frac{-\dfrac{z-\bar z}{2i}}{2\dfrac{z+\bar z}{2}\dfrac{z-\bar z}{2i} +1} = - \frac{z - \bar z}{z^2-\bar z^2+2i} \\[5px]
&\iff\;\; z^3 \bar z - z \bar z^3 + 2i z \bar z + z - \bar z = 0 \tag{*}\\[5px]
&\iff\;\; \bar z(z^3+iz-1)-z(\bar z^3 - i \bar z -1) = 0
\end{align}
$$
The latter holds true since the two factors in parenthesis are $0$ from the original equation and its complex conjugate.
$(*)$ For this step to hold as an equivalence, it also needs to be shown that $z^2-\bar z^2+2i \ne 0$ for values of $z$ which satisfy the original equation (which is fairly straightforward).
