The number of real roots of the equation $z^{3} + i z - 1 = 0$ is

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Solution

The correct option is C $0$
Let
$z = x + i y$
$z^{3} = x^{3} - 3 x y^{2} + i (3 x^{2} y - y^{3})$
Hence
$z^{3} + i z - 1 = 0$
$x^{3} - 3 x y^{2} + i (3 x^{2} y - y^{3}) + i x - y - 1 = 0$
$(x^{3} - 3 x y^{2} - y - 1) + i (3 x^{2} y - y^{3} + x) = 0$
Hence
$x^{3} - 3 x y^{2} - y - 1 = 0$ And
$3 x^{2} y - y^{3} + x = 0$
For real root, y has to be 0.
Substituting in the equations, we get
$x^{3} = 1$
$x = 1$ and $x = 0$ .
However, none of the above satisfies the given equation.
Hence number of real roots is 0.

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