Question

Show that the equation $a{z}^3+b{z}^2+\overline{b}z+\overline{a}=0$ has a root a, such that |a|=1.a,b,z, and a belong to the set of complex numbers.

Answer 1

$\begin{array}{c}wehaveaequation,\\ a{z}^3+b{z}^2+\overline{b}z+\overline{a}=0------\left(iii\right)\\ \overline{a}{\overline{z}}^3+\overline{b}{\overline{z}}^2+b\overline{z}+a=0------\left(i\right)\\ Now,\\ Dividingequ\left(i\right)with{\overline{z}}^3\\ \overline{a}+\frac{\overline{b}}{\overline{z}}+\frac{b}{{\overline{z}}^2}+\frac{a}{{\overline{z}}^3}=0\\ \Rightarrow \frac{a}{{\overline{z}}^3}+\frac{b}{{\overline{z}}^2}+\frac{\overline{b}}{\overline{z}}+\overline{a}=0------\left(ii\right)\\ \Rightarrow a{z}^3+b{z}^2+\overline{b}z+\overline{a}=0\\ z^3=\frac{1}{{\overline{z}}^3},{\left|z\right|}^6=1,\left|z\right|=1\\ \therefore \left|\alpha \right|=1\\ so,\alpha istherootoftheequis\left|\alpha \right|=1\end{array}$