Let four points z1, z2, z3, z4 be in complex plane such that z1<1, z2=1 and z3≤1. If z3=z2z1−z4z1z4−1, then |z4| can be

z3=z2z1−z4z¯1z4−1⇒ z3=z2z1−z4z1¯z4−1=z1−z4z1¯z4−1≤1If and only if z1−z42≤z1z4−12⇒ z12+z22−z1z4−z1z4≤z1z42+1−z1z4−z1z4⇒ z12+z42−z12z42−1≤0⇒ z12−1z42−1≥0Since z1≤1, we must have z4≤1

Let four points z1, z2, z3, z4 be in complex plane such that z1<1, z2=1 and z3≤1. If z3=z2z1−z4z1z4−1, then |z4| can be

z3=z2z1−z4z¯1z4−1⇒ z3=z2z1−z4z1¯z4−1=z1−z4z1¯z4−1≤1If and only if z1−z42≤z1z4−12⇒ z12+z22−z1z4−z1z4≤z1z42+1−z1z4−z1z4⇒ z12+z42−z12z42−1≤0⇒ z12−1z42−1≥0Since z1≤1, we must have z4≤1

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Let four points z₁, z₂, z₃, z₄ be in complex plane such that $|z_{1}| < 1, |z_{2}| = 1$ and $|z_{3}| \leq 1$ . If $z_{3} = \frac{z_{2} (z_{1} - z_{4})}{z_{1} z_{4} - 1}$ , then |z₄| can be

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answer is B, C.

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Detailed Solution

$\begin{matrix} z_{3} = \frac{z_{2} (z_{1} - z_{4})}{{\bar{z}}_{1} z_{4} - 1} \\ \Rightarrow |z_{3}| = \frac{|z_{2}| |z_{1} - z_{4}|}{|\bar{z_{1}} z_{4} - 1|} = \frac{|z_{1} - z_{4}|}{|\bar{z_{1}} z_{4} - 1|} \leq 1 \end{matrix}$

If and only if ${|z_{1} - z_{4}|}^{2} \leq {|z_{1} z_{4} - 1|}^{2}$

$\begin{array}{l} \Rightarrow {|z_{1}|}^{2} + {|z_{2}|}^{2} - z_{1} z_{4} - z_{1} z_{4} \leq {|z_{1} z_{4}|}^{2} + 1 - z_{1} z_{4} - z_{1} z_{4} \\ \Rightarrow {|z_{1}|}^{2} + {|z_{4}|}^{2} - {|z_{1}|}^{2} {|z_{4}|}^{2} - 1 \leq 0 \\ \Rightarrow ({|z_{1}|}^{2} - 1) ({|z_{4}|}^{2} - 1) \geq 0 \end{array}$