Question

Find the number of real or purely imaginary solution of the equation $z^3+iz-1=0$ is:

• A. $zero$
• B. one
• C. two
• D. three

Answer 1

The correct option is A $zero$

Let, $z=x+iy$ where $x,y\in R$ . Then,

$(x+iy{)}^3+i(x+iy)-1=0$

$⟹(x^3-3x{y}^2-y-1)+i(3x^{2}y-y^3+x)=0$

So, we have

$x^3-3x{y}^2-y-1=0=3x^{2}y-y^3+x$

If $y=0$ then $x^3-1=0=x$ there is no such $x\in R$

If $x=0$ then $-y-1=0=-y$ there is no such $y\in R$

So, Option A is correct.