Question

For any two complex numbers $Z_1$ and $Z_2$ and any real numbers $a$ and $b$, $\left|\right(a{z}_1–b{z}_2){|}^2+|(a{z}_1+b{z}_2){|}^2$ is equal to

• A. $(a^2+b^2)\left(\right|z_1|+|z_2\left|\right)$
• B. $(a^2+b^2)\left(\right|z_1{|}^2+\left|z_2{|}^2\right)$
• C. $(a^2+b^2){\left(\right|z_1|-|z_2\left|\right)}^2$
• D. None of these

Answer 1

The correct option is B

$(a^2+b^2)\left(\right|z_1{|}^2+\left|z_2{|}^2\right)$

Find the value of $\mathbf{\left|}\mathbf{\right(}\mathit{a}{\mathit{z}}_{\mathbf{1}}\mathbf{}\mathbf{–}\mathbf{}\mathit{b}{\mathit{z}}_{\mathbf{2}}\mathbf{)}{\mathbf{|}}^{\mathbf{2}}\mathbf{+}\mathbf{}\mathbf{|}\mathbf{(}\mathit{a}{\mathit{z}}_{\mathbf{1}}\mathbf{}\mathbf{+}\mathbf{}\mathit{b}{\mathit{z}}_{\mathbf{2}}\mathbf{)}{\mathbf{|}}^{\mathbf{2}}\mathbf{:}$:

Given that, $\left|\right(a{z}_1–b{z}_2){|}^2+|(a{z}_1+b{z}_2){|}^2$

$=\left(a{z}_1-b{z}_2\right)\left(\overline{a{z}_1-b{z}_2}\right)+\left(a{z}_1+b{z}_2\right)\left(\overline{a{z}_1+b{z}_2}\right)$ $\left[\mathbf{\because }\mathit{z}\mathbf{·}\overline{\mathbf{z}}\mathbf{}\mathbf{=}\mathbf{}{\left|z\right|}^{\mathbf{2}}\right]$

$=\left(a{z}_1-b{z}_2\right)\left(a\overline{z_1}-b\overline{z_2}\right)+\left(a{z}_1+b{z}_2\right)\left(a\overline{z_1}+b\overline{z_2}\right)$

$=a^2{\left|z_1\right|}^2-ab{z}_1\overline{z_2}-ab{z}_2\overline{z_1}+b^2{\left|z_2\right|}^2+b^2{\left|z_1\right|}^2+ab{z}_1\overline{z_2}+ab{z}_2\overline{z_2}+a^2{\left|z_2\right|}^2$

$=a^2{\left|z_1\right|}^2+b^2{\left|z_2\right|}^2+b^2{\left|z_1\right|}^2+a^2{\left|z_2\right|}^2$

$={\left|z_1\right|}^2\left(a^2+b^2\right)+{\left|z_2\right|}^2\left(a^2+b^2\right)$

$\mathbf{\therefore }\mathbf{\left|}\mathbf{\right(}\mathit{a}{\mathit{z}}_{\mathbf{1}}\mathbf{}\mathbf{–}\mathbf{}\mathit{b}{\mathit{z}}_{\mathbf{2}}\mathbf{)}{\mathbf{|}}^{\mathbf{2}}\mathbf{+}\mathbf{}\mathbf{|}\mathbf{(}\mathit{a}{\mathit{z}}_{\mathbf{1}}\mathbf{}\mathbf{+}\mathbf{}\mathit{b}{\mathit{z}}_{\mathbf{2}}\mathbf{)}{\mathbf{|}}^{\mathbf{2}}$ $=\mathbf{(}{\mathit{a}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{b}}^{\mathbf{2}}\mathbf{)}\left({\left|z_1\right|}^2+{\left|z_2\right|}^2\right)$

Hence, Option ‘B’ is Correct.