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Mathematics Question on Complex Numbers and Quadratic Equations

If z1z_1 and z2z_2 are any two complex numbers, then ∣z1+z12−z22∣+∣z1+z12−z22∣|z_1+\sqrt {z_1^2-z_2^2}|+ |z_1+\sqrt {z_1^2-z_2^2}| is equal to

A

∣z1∣|z_1|

B

∣z2∣|z_2|

C

∣z1+z2∣|z_1+z_2|

D

none of these

Answer

∣z1+z2∣|z_1+z_2|

Explanation

Solution

We know that ∣z1+z2∣2∣z1−z2∣2\left|z_{1}+z_{2}\right|^{2}\left|z_{1}-z_{2}\right|^{2} =2[∣z1∣2+∣z2∣2…(1)=2[\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\quad\ldots\left(1\right) Now [z1+z12−z22+∣z1−z12−z22∣]2\left[z_{1}+\sqrt{z^{2}_{1}-z^{2}_{2}}+\left|z_{1}-\sqrt{z^{2}_{1}-z^{2}_{2}}\right|\right]^{2} =∣z1+z12−z22∣2+∣z1−z12−z22∣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∣z12−(z12−z22)∣+2\left|z^{2}_{1}-\left(z^{2}_{1}-z^{2}_{2}\right)\right| =2∣z1∣2+2∣z12−z22∣+2∣z22∣=2\left|z_{1}\right|^{2}+2\left|z_{1}^{2}-z^{2}_{2}\right|+2\left|z^{2}_{2}\right|\quad [By (1)(1)] =2∣z1∣2+2∣z2∣2+2∣z12−z22∣=2|z_1|^2+2|z_2|^2+2|z^2_1-z^2_2| =∣z1+z2∣2+∣z1−z2∣2+2∣z1+z2∣∣z1−z2∣=|z_1+z_2|^2+|z_1-z_2|^2+2|z_1+z_2| |z_1-z_2| =(∣z1+z2∣+∣z1−z2∣)2= (|z_1 + z_2| + |z_1 - z_2|)^2 Taking square root of both sides, we get ∣z1+z12−z22∣+∣z1−z12−z22∣\left|z_{1}+\sqrt{z^{2}_{1}-z^{2}_{2}}\right|+\left|z_{1}-\sqrt{z^{2}_{1}-z^{2}_{2}}\right| =∣z1+z2∣+∣z1−z2∣=\left|z_{1}+z_{2}\right|+\left|z_{1}-z_{2}\right|