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

The Correct Option is C

We know that $\left|z_{1}+z_{2}\right|^{2}\left|z_{1}-z_{2}\right|^{2}$ $=2[\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\quad\ldots\left(1\right)$ Now $\left[z_{1}+\sqrt{z^{2}_{1}-z^{2}_{2}}+\left|z_{1}-\sqrt{z^{2}_{1}-z^{2}_{2}}\right|\right]^{2}$ $=\left|z_{1}+\sqrt{z^{2}_{1}-z^{2}_{2}}\right|^{2}+\left|z_{1}-\sqrt{z^{2}_{1}-z^{2}_{2}}\right|^{2}$ $+2\left|z^{2}_{1}-\left(z^{2}_{1}-z^{2}_{2}\right)\right|$ $=2\left|z_{1}\right|^{2}+2\left|z_{1}^{2}-z^{2}_{2}\right|+2\left|z^{2}_{2}\right|\quad$ [By $(1)$] $=2|z_1|^2+2|z_2|^2+2|z^2_1-z^2_2|$ $=|z_1+z_2|^2+|z_1-z_2|^2+2|z_1+z_2| |z_1-z_2|$ $= (|z_1 + z_2| + |z_1 - z_2|)^2$ Taking square root of both sides, we get $\left|z_{1}+\sqrt{z^{2}_{1}-z^{2}_{2}}\right|+\left|z_{1}-\sqrt{z^{2}_{1}-z^{2}_{2}}\right|$ $=\left|z_{1}+z_{2}\right|+\left|z_{1}-z_{2}\right|$