The equation ${\mathrm{az}}^3+{\mathrm{bz}}^2+\overline{\mathrm{b}}\mathrm{z}+\overline{\mathrm{a}}=0$ has a root $\mathrm{\alpha }$; where a, b, z and $\mathrm{\alpha }$ belong to the set of complex numbers. The value of |$\mathrm{\alpha }$| is

We have ${\mathrm{az}}^3+{\mathrm{bz}}^2+\overline{\mathrm{b}}\mathrm{z}+\overline{\mathrm{a}}=0$            ……. (1)

Taking conjugate of both sides, we get

$\Rightarrow \overline{\mathrm{a}} {\overline{\mathrm{z}}}^3+\overline{\mathrm{b}} {\overline{\mathrm{z}}}^2+\mathrm{b}\overline{\mathrm{z}}+\mathrm{a}=0$

Dividing this equation by ${\overline{z}}^3$ and writing the terms in reverse order we get

$\frac{\mathrm{a}}{{\overline{\mathrm{z}}}^3}+\frac{\mathrm{b}}{{{\displaystyle \overline{\mathrm{z}}}}^2}+\frac{\overline{\mathrm{b}}}{\overline{\mathrm{z}}}+\overline{\mathrm{a}}=0$             [For $\overline{\mathrm{z}}\ne 0$]     …… (2)

Since equations (1) and (2) are identical,

${\mathrm{z}}^3=\frac{1}{{\overline{\mathrm{z}}}^3}\Rightarrow |\mathrm{z}{|}^6=1\Rightarrow |\mathrm{z}|=1$

Therefore, if $\mathrm{\alpha }$ is a root of the given equation, then $\left|\mathrm{\alpha }\right|=1$.