Algebra – Complex Numbers – JEE Main 21 Jan 2026 Shift 2

Solution:

The region $|z-5|\le3$ represents a circle with center $C(5,0)$ and radius $r=3$.

For $z$ to have the maximum positive principal argument, the corresponding point $P$ must lie on the tangent drawn from the origin to the circle in the first quadrant.

Let $\angle POC = \theta$. In $\triangle OPC$:
$$\sin \theta = \frac{PC}{OC} = \frac{3}{5}$$
$$\cos \theta = \sqrt{1 – \left(\frac{3}{5}\right)^2} = \frac{4}{5}$$

The distance $OP = |z|$ is:
$$|z| = \sqrt{OC^2 – PC^2} = \sqrt{5^2 – 3^2} = \sqrt{16} = 4$$

So, the complex number $z$ is:
$$z = |z|(\cos \theta + i \sin \theta)$$
$$z = 4\left(\frac{4}{5} + i\frac{3}{5}\right) = \frac{16}{5} + i\frac{12}{5}$$
$$\Rightarrow 5z = 16 + 12i$$

Now evaluate the term inside the modulus:
$$\frac{5z-12}{5iz+16} = \frac{(16+12i)-12}{i(16+12i)+16}$$
$$= \frac{4+12i}{16i-12+16} = \frac{4+12i}{4+16i}$$
$$= \frac{4(1+3i)}{4(1+4i)} = \frac{1+3i}{1+4i}$$

Now substitute this back into the expression:
$$34\left|\frac{5z-12}{5iz+16}\right|^{2} = 34\left|\frac{1+3i}{1+4i}\right|^{2}$$
$$= 34 \left( \frac{|1+3i|^2}{|1+4i|^2} \right)$$
$$= 34 \left( \frac{1^2 + 3^2}{1^2 + 4^2} \right)$$
$$= 34 \left( \frac{10}{17} \right)$$
$$= 2 \times 10 = 20$$

Ans. (4)