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

$$\begin{gathered} {\mathrm{z}}^3+\mathrm{iz}-1=0 \\ \mathrm{Let} \mathrm{z}=\mathrm{\alpha } \mathrm{be} \mathrm{the} \mathrm{real} \mathrm{root} \\ \Rightarrow \left({\mathrm{\alpha }}^3-1\right)+\mathrm{i\alpha }=0 \\ \Rightarrow {\mathrm{\alpha }}^3-1=0 \mathrm{and} \mathrm{\alpha }=0 \end{gathered}$$

which is not possible $\Rightarrow$the equation has no real root

$\therefore$no. of real roots=0