Question

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, $(x+iy{)}^3$ $+i(x+iy)—1=0$

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

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

So, the number of real or imaginary solutions of the equation is 0.