Let z be the complex number satisfying |z-5| ≤ 3 and having maximum positive principal argument. Then 34 | 5 z-12 5 i z+16 |^2 is equal to :
JEE Main 2026 — Mathematics Algebra
2026mcqmedium
Let z be the complex number satisfying ∣z−5∣≤3 and having maximum positive principal argument. Then 345iz+165z−122 is equal to :
Official previous-year question
Held on 21 Jan 2026 · Verified 6 Jul 2026.
Options
A
26
B
12
C
20
D
16
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Solution
The region ∣z−5∣≤3 is a disk centered at (5,0) with radius 3. Maximum positive principal argument from origin occurs at the tangent point from origin to the circle.
sinθ=53, cosθ=54.
The tangent point is the foot of perpendicular from (5,0) to the line 3x−4y=0, giving z=516+512i.
5z−12=4+12i⇒∣5z−12∣2=160
5iz+16=4+16i⇒∣5iz+16∣2=272
345iz+165z−122=34×272160=34×1710=20.
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