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

z 1 z 2 is purely real , arg ⁡ z 1 − arg ⁡ z 2 = π

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

z 1 2 − z 2 2 = z 1 2 + z 1 2 − 2 z 1 z 2 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 ⇒ z 1 z 2 + 1 = z 1 z 2 − 1 ⇒ z 1 z 2 lies on ⊥ bisector of 1 and -1 ⇒ z 1 z 2 lies on imaginary axis ⇒ z 1 z 2 is purely imaginary ⇒ arg ⁡ z 1 z 2 = ± π 2 arg ⁡ z 1 − arg ⁡ z 2 = π 2

arg ⁡ z 1 − arg ⁡ z 2 = π 2

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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

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

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