Question

If $z_1$ and $z_2$ are complex numbers and $u=\sqrt{z_1{z}_2},$ then $|\frac{z_1+z_2}{2}+u|+|\frac{z_1+z_2}{2}-u|$ is equal to

• A. $|z_1|-|z_2|$
• B. $|z_1{|}^2+|z_2{|}^2$
• C. $0$
• D. $|z_1|+|z_2|$

Answer 1

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

$|\frac{z_1+z_2}{2}+u|+|\frac{z_1+z_2}{2}-u|=|\frac{z_1+z_2}{2}+\sqrt{z_1{z}_2}|+|\frac{z_1+z_2}{2}-\sqrt{z_1{z}_2}|=\left|\frac{(\sqrt{z_1}+\sqrt{z_2}{)}^2}{2}\right|+\left|\frac{(\sqrt{z_1}-\sqrt{z_2}{)}^2}{2}\right|=\left|\frac{(p+q{)}^2}{2}\right|+\left|\frac{(p-q{)}^2}{2}\right|$
Where $p=\sqrt{z_1}$ and $q=\sqrt{z_2}$
$=\frac{1}{2}[|p+q{|}^2+|p-q{|}^2]=\frac{1}{2}[2|p{|}^2+2|q{|}^2]=|p^2|+|q^2|=|z_1|+|z_2|$


Alternate Solution :
Put $z_1=0,$ then expression becomes $|z_2|$
Put $z_2=0,$ then expression becomes $|z_1|$
$\therefore$ From options, $|z_1|+|z_2|$ is the CORRECT answer.