The real or purely imaginary solutions of the equation, $z^3+iz-1=0$ are? 
The number of real or purely imaginary solutions of the equation, $z^3+iz-1=0$
  is?

I substituted $x+iy=z$ and am getting two equations which seem impossible to solved.What would be the correct approach ?
 A: Let $z=x+iy$ where $x,y\in\mathbb R$. Then, 
$$(x+iy)^3+i(x+iy)-1=0,$$
i.e.
$$(x^3-3xy^2-y-1)+i(3x^2y-y^3+x)=0$$
So, we have
$$x^3-3xy^2-y-1=0=3x^2y-y^3+x$$
I guess that you already got this. You don't need to solve the system because the question says "the number of real or purely imaginary solutions".
If $y=0$, then $x^3-1=0=x$. There is no such $x\in\mathbb R$.
If $x=0$, then $-y-1=0=-y^3$. There is no such $y\in\mathbb R$.
A: If $z = \lambda i$ is a purely imaginary solution to this equation, then you end up with
$$
0 = (i \lambda)^3 + i (i \lambda) - 1 = -i\lambda^3 - \lambda - 1
$$
Note that since $\lambda$ is a real number in this case, we have that
$$
0 + i0 = (-\lambda - 1) + i(-\lambda)
$$
or equivalently, that $\lambda = 0$ and $- \lambda - 1 = 0$ i.e. $\lambda = 0$ and $\lambda = -1$. This is a contradiction, so there are no purely imaginary solutions.
The count of purely real solutions should proceed similarly.
A: The question isn't asking about solutions of the form $z=x+iy$, so you're making things much harder on yourself than you need to.
Rather, it asks about two cases: $ z=x$ (where $x$ is real) and $z=iy$ (where $y$ is real). 
Consider those two simpler cases separately and you might make better progress.
