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

Question

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 ?

Answer 1

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 $y\in\mathbb R$.

Answer 2

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.

Answer 3

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.