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 :

Correct answer: C. 20 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. θ = (3)/(5) , θ = (4)/(5) . The tangent point is the foot of perpendicular from (5,0) to the line 3x - 4y = 0 , giving z = (16)/(5) + (12)/(5)i . 5z - 12 = 4 + 12i |5z - 12|^2 = 160 5iz + 16 = 4 + 16i |5iz + 16|^2 = 272 34|(5z-12)/(5iz+16)|^2 = 34 × (160)/(272) = 34 × (10)/(17) = 20 .

JEE Main 202621 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 21 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2026 · 21 January · Evening Shift · Mathematics · Complex Number

Let z be the complex number satisfying |z-5| ≤ 3 and having maximum...

Let

z

be the complex number satisfying

|z-5| \leq 3

and having maximum positive principal argument. Then

34\left|\frac{5 z-12}{5 \mathrm{i} z+16}\right|^{2}

is equal to :

Options

Correct answer

C. 20

Step-by-step solution

The region

|z - 5| \le 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.

\sin\theta = \frac{3}{5}

\cos\theta = \frac{4}{5}

.
The tangent point is the foot of perpendicular from

(5,0)

to the line

3x - 4y = 0

, giving

z = \frac{16}{5} + \frac{12}{5}i

5z - 12 = 4 + 12i \Rightarrow |5z - 12|^2 = 160

5iz + 16 = 4 + 16i \Rightarrow |5iz + 16|^2 = 272

34\left|\frac{5z-12}{5iz+16}\right|^2 = 34 \times \frac{160}{272} = 34 \times \frac{10}{17} = 20

Practice Complex Number on Quantrex Academy →

More from Complex Number