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. $|z_1+z_2|$
• B. $\left|z_1\right|$
• C. $\left|z_2\right|$
• D. None of these

Answer 1

The correct option is D None of these

We know that ${|z_1+z_2|}^2+{|z_1-z_2|}^2=2[{\left|z_1\right|}^2+{\left|z_2\right|}^2]$ ...(1)

Now, ${[|z_1+\sqrt{{z_1}^2-{z_2}^2}|+|z_1-\sqrt{{z_1}^2-{z_2}^2}|]}^2$

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

$=2{\left|z_1\right|}^2+2|{z_1}^2-{z_2}^2|+2\left|{z_2}^2\right|$ by (1)

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

Taking square root of both sides, we get

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