Question

If $z_1$ and $z_2$ are any two complex numbers, then $|z_1+\sqrt{z_1^2-z_2^2}|+|z_1+\sqrt{z_1^2-z_2^2}|$ is equal to

• A. $|z_1|$
• B. $|z_2|$
• C. $|z_1+z_2|$
• D. none of these

Answer 1

Answer: (C)

We know that
${\left|z_1+z_2\right|}^2{\left|z_1-z_2\right|}^2$ $=2[{\left|z_1\right|}^2+{\left|z_2\right|}^2\dots \left(1\right)$
Now ${\left[z_1+\sqrt{z_1^2-z_2^2}+\left|z_1-\sqrt{z_1^2-z_2^2}\right|\right]}^2$
$={\left|z_1+\sqrt{z_1^2-z_2^2}\right|}^2+{\left|z_1-\sqrt{z_1^2-z_2^2}\right|}^2$
$+2\left|z_1^2-\left(z_1^2-z_2^2\right)\right|$
$=2{\left|z_1\right|}^2+2\left|z_1^2-z_2^2\right|+2\left|z_2^2\right|$ [By $(1)$]
$=2|z_1{|}^2+2|z_2{|}^2+2|z_1^2-z_2^2|$
$=|z_1+z_2{|}^2+|z_1-z_2{|}^2+2|z_1+z_2||z_1-z_2|$
$=(|z_1+z_2|+|z_1-z_2|{)}^2$
Taking square root of both sides, we get
$\left|z_1+\sqrt{z_1^2-z_2^2}\right|+\left|z_1-\sqrt{z_1^2-z_2^2}\right|$
$=\left|z_1+z_2\right|+\left|z_1-z_2\right|$