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Show that the locus formed by $z$ in the equation $z^3+iz=1$ never crosses the coordinate axes in the Argand's plane.Further show that $|z|=\sqrt{\frac{-\text{Im}(z)}{2\text{Re}(z)\text{Im}(z)+1}}$


Let $z=x+iy$.Then $z^3+iz=1$ becomes $(x+iy)^3+i(x+iy)=1$

$x^3-iy^3+3ix^2y-3xy^2+ix-y=1$

$(x^3-3xy^2-y)+i(3x^2y-y^3+x)=1$

Comparing real and imaginary parts on both sides,we get
$x^3-3xy^2-y=1,3x^2y-y^3+x=0$

But i do not know how to prove that the locus formed by $z$ never crosses the coordinate axes in the Argand's plane.

In the second part,$|z|=\sqrt{x^2+y^2}$ and $\sqrt{\frac{-\text{Im}(z)}{2\text{Re}(z)\text{Im}(z)+1}}=\sqrt{\frac{-y}{2xy+1}}$.I don't understand how to prove them equal.

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Note that the problem is worded somewhat misleadingly -- the "locus" it asks about is not a curve (as the word "crosses" would also appear to imply), but consists of just three isolated points, the roots of the polynomial $z^3+iz-1$.

For the first part, I would just say:

  1. Clearly $0$ is not a solution.

  2. If $z$ is nonzero real, then $z^3$ is also real, and $iz$ is imaginary. Thus the left-hand side of the equation cannot be real.

  3. If $z$ is nonzero imaginary, then $iz$ is real, but now $z^3$ is nonzero imaginary, and the left-hand side of the equation still cannot be real.

For the second part, I don't see anything particularly clever to do. Just square both sides of the equation you're trying to prove, cross-multiply by the denominator, and verify that what you get is implied by the real and imaginary parts of $z^3+iz=1$ which you know holds.

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