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Maximum of |2z1 - 3z2 - 4z3| for Three Circles | JEE Advanced

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

Question
If ∣z1∣=1\left|z_1\right| = 1, ∣z2−2∣=3\left|z_2 - 2\right| = 3 and ∣z3−5∣=6\left|z_3 - 5\right| = 6, then the maximum value of ∣2z1−3z2−4z3∣\left|2z_1 - 3z_2 - 4z_3\right| is not less than (where z1,z2,z3z_1, z_2, z_3 are complex numbers)
A4343correct
B4444correct
C4545correct
D4646correct
Solution
Step 1: Notice that the data controls z1z_1, z2−2z_2 - 2 and z3−5z_3 - 5, not z2z_2 and z3z_3 themselves. So rewrite the expression in those pieces.
−3z2=−3(z2−2)−6,−4z3=−4(z3−5)−20.-3z_2 = -3\left(z_2 - 2\right) - 6, \qquad -4z_3 = -4\left(z_3-5\right) - 20 .
 Step 2: Substitute.
2z1−3z2−4z3=2z1−3(z2−2)−4(z3−5)−26.2z_1 - 3z_2 - 4z_3 = 2z_1 - 3\left(z_2-2\right) - 4\left(z_3-5\right) - 26 .
 Step 3: Apply the triangle inequality ∣A+B+C+D∣≤∣A∣+∣B∣+∣C∣+∣D∣|A+B+C+D| \le |A|+|B|+|C|+|D|.
∣2z1−3z2−4z3∣≤2∣z1∣+3∣z2−2∣+4∣z3−5∣+26.\left|2z_1 - 3z_2 - 4z_3\right| \le 2\left|z_1\right| + 3\left|z_2-2\right| + 4\left|z_3-5\right| + 26 .
 Step 4: Put in the given moduli.
≤2(1)+3(3)+4(6)+26=2+9+24+26=61.\le 2(1) + 3(3) + 4(6) + 26 = 2 + 9 + 24 + 26 = 61 .
 Step 5: Check the bound is reachable. Equality in the triangle inequality needs all four terms to point the same way, which we can arrange: each of z1z_1, z2−2z_2 - 2, z3−5z_3 - 5 is free to take any direction on its circle. So the maximum really is 6161. Step 6: Compare with the options. Since 61≥43, 44, 45, 4661 \ge 43,\ 44,\ 45,\ 46, the maximum is not less than any of them. Answer: (1), (2), (3) and (4).
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