Q. For any two complex numbers $\mathcal{z}_{1}$ and $\mathcal{z}_{2}$ and any real numbers $a$ and $b;\left| a\mathcal{z}_{1} - b\mathcal{z}_{2} \right|^{2} + |\left( b\mathcal{z}_{1} + a\mathcal{z}_{2} \right)\left. \ \right|^{2}$ is equal to
A) $\left( a^{2} + b^{2} \right)(\left| \mathcal{z}_{1} \right| + \left| \mathcal{z}_{2} \right|)$
B) $\left( a^{2} + b^{2} \right)(\left| \mathcal{z}_{1} \right|^{2} + \left| \mathcal{z}_{2} \right|^{2})$
C) $\left( a^{2} + b^{2} \right)(\left| \mathcal{z}_{1} \right|^{2} - \left| \mathcal{z}_{2} \right|^{2})$
D) None of the above
Explanation
(b)
Now, $\left| (a\mathcal{z}_{1} - b\mathcal{z}_{2} \right)\left. \ \right|^{2} + |\left( b\mathcal{z}_{1} + a\mathcal{z}_{2} \right)\left. \ \right|^{2}$
$= a^{2}\left| \mathcal{z}_{1} \right|^{2} + b^{2}\left| \mathcal{z}_{2} \right|^{2} - 2ab\ Re\left| \mathcal{z}_{1}{\overline{\mathcal{z}}}_{2} \right| + b^{2}\left| \mathcal{z}_{1} \right|^{2} + a^{2}\left| \mathcal{z}_{2} \right|^{2} + 2ab\ Re\left| {\overline{\mathcal{z}}}_{1}\mathcal{z}_{2} \right|$
$= \left( a^{2} + b^{2} \right)(\left| \mathcal{z}_{1} \right|^{2} + \left| \mathcal{z}_{2} \right|^{2})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.