Q. If $\mathcal{z}$ is a complex number, then $\left( {\overline{\mathcal{z}}}^{- 1} \right)(\overline{\mathcal{z}})$ is equal to
A) 1
B) $- 1$
C) 0
D) None of these
Explanation
(a)
Let $\mathcal{z =}x + iy$
$\Rightarrow \ \ \overline{\mathcal{z}} = x - iy$
and $\left( {\overline{\mathcal{z}}}^{- 1} \right) = \frac{1}{x - iy} = \frac{x + iy}{x^{2} + y^{2}}$
$\therefore\ \ \left( {\overline{\mathcal{z}}}^{- 1} \right)\overline{\mathcal{z}} = \frac{x + iy}{x^{2} + y^{2}} \times (x - iy) = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.