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Bounding |z4| in the Blaschke Relation z3 = z2(z1-z4)/(conj(z1)z4-1) | JEE

JEE Maths question with a full step-by-step solution.

Question
Given z1,z2,z3,z4z_1, z_2, z_3, z_4 are four points in the complex plane such that ∣z1∣<1\left|z_1\right| < 1, ∣z2∣=1\left|z_2\right| = 1 and ∣z3∣≤1\left|z_3\right| \le 1, and
z3=z2(z1−z4)z1ˉz4−1,z_3 = \frac{z_2\left(z_1 - z_4\right)}{\bar{z_1}z_4 - 1} ,
 then ∣z4∣\left|z_4\right| can be (z1≠z4)\left(z_1 \ne z_4\right)
A22
B25\dfrac25correct
C13\dfrac13correct
D52\dfrac52
Solution
Step 1: Take the modulus of the given relation and use ∣z2∣=1\left|z_2\right| = 1.
∣z3∣=∣z2∣∣z1−z4∣∣z1ˉz4−1∣=∣z1−z4∣∣z1ˉz4−1∣.\left|z_3\right| = \frac{\left|z_2\right|\left|z_1-z_4\right|}{\left|\bar{z_1}z_4-1\right|} = \frac{\left|z_1-z_4\right|}{\left|\bar{z_1}z_4-1\right|} .
 Step 2: Impose ∣z3∣≤1\left|z_3\right| \le 1 and clear the denominator.
∣z1−z4∣≤∣z1ˉz4−1∣.\left|z_1 - z_4\right| \le \left|\bar{z_1}z_4 - 1\right| .
 Step 3: Square both sides, using ∣A∣2=AAˉ|A|^{2} = A\bar A.
(z1−z4)(z1ˉ−z4ˉ)≤(z1ˉz4−1)(z1z4ˉ−1).\left(z_1-z_4\right)\left(\bar{z_1}-\bar{z_4}\right) \le \left(\bar{z_1}z_4-1\right)\left(z_1\bar{z_4}-1\right).
 Step 4: Expand the left side.
∣z1∣2+∣z4∣2−z1z4ˉ−z1ˉz4.\left|z_1\right|^{2} + \left|z_4\right|^{2} - z_1\bar{z_4} - \bar{z_1}z_4 .
 Step 5: Expand the right side.
∣z1∣2∣z4∣2−z1ˉz4−z1z4ˉ+1.\left|z_1\right|^{2}\left|z_4\right|^{2} - \bar{z_1}z_4 - z_1\bar{z_4} + 1 .
 Step 6: The two cross terms are identical on both sides, so they cancel:
∣z1∣2+∣z4∣2≤∣z1∣2∣z4∣2+1.\left|z_1\right|^{2} + \left|z_4\right|^{2} \le \left|z_1\right|^{2}\left|z_4\right|^{2} + 1 .
 Step 7: Move everything to one side and factorise.
∣z1∣2−∣z1∣2∣z4∣2+∣z4∣2−1≤0\left|z_1\right|^{2} - \left|z_1\right|^{2}\left|z_4\right|^{2} + \left|z_4\right|^{2} - 1 \le 0
  ⟹  ∣z1∣2(1−∣z4∣2)−(1−∣z4∣2)≤0\implies \left|z_1\right|^{2}\left(1 - \left|z_4\right|^{2}\right) - \left(1 - \left|z_4\right|^{2}\right) \le 0
  ⟹  (∣z1∣2−1)(1−∣z4∣2)≤0.\implies \left(\left|z_1\right|^{2} - 1\right)\left(1 - \left|z_4\right|^{2}\right) \le 0 .
 Step 8: Use ∣z1∣<1\left|z_1\right| < 1, which makes the first bracket strictly negative. Dividing by a negative number flips the inequality:
1−∣z4∣2≥0  ⟹  ∣z4∣≤1.1 - \left|z_4\right|^{2} \ge 0 \implies \left|z_4\right| \le 1 .
 Step 9: Check the options against ∣z4∣≤1\left|z_4\right| \le 1: 25\dfrac25 and 13\dfrac13 qualify, while 22 and 52\dfrac52 do not. Answer: (2) and (3).
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