JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Complex Numbers

Let z be the complex number satisfying

∣ z - 5∣ \leq 3

and having maximum positive principal argument.
Then

34 \frac{5 z - 12}{5 i z + 16}^{2}

is equal to:

(A)16
(B)12
(C)26
(D)20✓ Correct

Answer: (D)

Solution

The condition

∣ z - 5∣ \leq 3

describes a disk centered at 5 with radius 3. For maximum positive argument, z lies on the circle at the point where the tangent from origin touches the circle.

Using geometry, this point is at angle

θ

where

sin θ = \frac{3}{5}

cos θ = \frac{4}{5}

.

The point is at distance 4 from origin:

z = 4 cos θ + 4 i sin θ = \frac{16}{5} + \frac{12 i}{5}

Now compute:

34 \frac{5 z - 12}{5 i z + 16}^{2} = 34 \frac{16 + 12 i - 12}{16 i - 12 + 16}^{2} = 34 \frac{4 + 12 i}{4 + 16 i}^{2}

= 34 \cdot \frac{16 + 144}{16 + 256} = 34 \cdot \frac{160}{272} = 20

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Complex Numbers

Solution

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