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. $|z_1+z_2|+|z_1-z_2|$

Answer 1

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

Consider ${\{|z_1+\sqrt{z_1^2-z_2^2}|+|z_1-\sqrt{z_1^2-z_2^2}|\}}^2$
$={|z_1+\sqrt{z_1^2-z_2^2}|}^2+{|z_1-\sqrt{z_1^2-z_2^2}|}^2+2|z_1+\sqrt{z_1^2-z_2^2}||z_1-\sqrt{z_1^2-z_2^2}|$

Using $|z_1+z_2{|}^2+|z_1-z_2{|}^2=2|z_1{|}^2+2|z_2{|}^2$ and $|z_1{z}_2|=|z_1||z_2|,$ given expression is equal to
$2|z_1{|}^2+2|z_1^2-z_2^2|+2|z_1^2-(z_1^2-z_2^2)|$
$=2|z_1{|}^2+2|z_1-z_2||z_1+z_2|+2|z_2^2|$
$=2|z_1^2|+2|z_2^2|+2|z_1-z_2||z_1+z_2|$

Using again $|z_1+z_2{|}^2+|z_1-z_2{|}^2=2|z_1{|}^2+2|z_2{|}^2,$ the expression becomes
$|z_1+z_2{|}^2+|z_1-z_2{|}^2+2|z_1+z_2||z_1-z_2|$
$={\{|z_1+z_2|+|z_1-z_2|\}}^2$
So, ${\{|z_1+\sqrt{z_1^2-z_2^2}|+|z_1-\sqrt{z_1^2-z_2^2}|\}}^2={\{|z_1+z_2|+|z_1-z_2|\}}^2$
Taking square root both sides, we get
$|z_1+\sqrt{z_1^2-z_2^2}|+|z_1-\sqrt{z_1^2-z_2^2}|=|z_1+z_2|+|z_1-z_2|$

Alternate :
Put $z_1=a,a>0$ and $z_2=0$ in the expression,
$|z_1+\sqrt{z_1^2-z_2^2}|+|z_1-\sqrt{z_1^2-z_2^2}|=2a$
Also, $|z_1+z_2|+|z_1-z_2|=2a$