Question

lf $z_1,z_2$ are any two complex numbers then $\left|z_{1^{+}}\sqrt{z_1^2-z_2^2}\right|+|z_1-\sqrt{z_1^2-z_2^2}|$ is equal to

• A. $|z_1|$
• B. $|z_2|$
• C. $|z_1+z_2|+|z_1-z_2|$
• D. $|z_1+z_2|-|z_1-z_2|$

Answer 1

The correct option is C $|z_1+z_2|+|z_1-z_2|$

${(|z_1+\sqrt{z_1^2-z_2^2}|+|z_1-\sqrt{z_1^2-z_2^2}|)}^2$

$={\left|z_1\right|}^2+{\left|z_1\right|}^2-{\left|z_2\right|}^2+{\left|z_1\right|}^2+{\left|z_1\right|}^2-{\left|z_2\right|}^2+2\left|z_1\right|\left|\sqrt{z_1^2-z_2^2}\right|+(-2\left|z_1\right|\left|\sqrt{z_1^2-z_2^2}\right|)+2({\left|z_1\right|}^2-|{\left|z_1\right|}^2-{\left|z_2\right|}^2|)$

$=4z_1^2⟶\left(1\right)$ (On simplifying )

$\Rightarrow {(|z_1+z_2|+|z_1-z_2|)}^2={\left|z_1\right|}^2+{\left|z_2\right|}^2+2\left|z_1\right|\left|z_2\right|+{\left|z_1\right|}^2-2\left|z_1\right|\left|z_2\right|+{({\left|z_1\right|}^2-{\left|z_2\right|}^2)}^2$

$=4z_1^2⟶\left(2\right)$ (On Simplifying )

$\Rightarrow \left(1\right)=\left(2\right)$

Hence, the answer is $|z_1+z_2|+|z_1-z_2|.$