If z1, z2 are two complex numbers z1≠z2 satisfying, z21−z22=z12+z22−2z1z2, then

z12−z22=z12+z12−2z1z2or z1−z2z1+z2=z1−z22or z1+z2=z1−z2 z1+z2=z1−z2⇒ z1z2+1=z1z2−1⇒ z1z2 lies on ⊥ bisector of 1 and -1⇒ z1z2 lies on imaginary axis⇒ z1z2 is purely imaginary⇒ arg⁡z1z2=±π2arg⁡z1−arg⁡z2=π2

If z1, z2 are two complex numbers z1≠z2 satisfying, z21−z22=z12+z22−2z1z2, then

z12−z22=z12+z12−2z1z2or z1−z2z1+z2=z1−z22or z1+z2=z1−z2 z1+z2=z1−z2⇒ z1z2+1=z1z2−1⇒ z1z2 lies on ⊥ bisector of 1 and -1⇒ z1z2 lies on imaginary axis⇒ z1z2 is purely imaginary⇒ arg⁡z1z2=±π2arg⁡z1−arg⁡z2=π2

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If z₁, z₂ are two complex numbers $(z_{1} \neq z_{2})$ satisfying, $|{z^{2}}_{1} - z_{2}^{2}| = |z_{1}^{2} + z_{2}^{2} - 2 z_{1} z_{2}|$ , then

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$\frac{z_{1}}{z_{2}}$ is purely real

$\frac{z_{1}}{z_{2}}$ is purely imaginary

$|\arg z_{1} - \arg z_{2}| = \frac{π}{2}$

$|\arg z_{1} - \arg z_{2}| = π$

answer is A, D.

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Detailed Solution

$|z_{1}^{2} - z_{2}^{2}| = |z_{1}^{2} + z_{1}^{2} - 2 z_{1} z_{2}|$

$\begin{array}{l} or |z_{1} - z_{2}| |z_{1} + z_{2}| = {|z_{1} - z_{2}|}^{2} \\ or |z_{1} + z_{2}| = |z_{1} - z_{2}| \\ |z_{1} + z_{2}| = |z_{1} - z_{2}| \\ \Rightarrow |\frac{z_{1}}{z_{2}} + 1| = |\frac{z_{1}}{z_{2}} - 1| \end{array}$