JEE Mathematics 2026 Question with Solution

Given: $|z - 5| \le 3$ represents a disk of radius $3$ centered at $(5,0)$ in the complex plane.

Find: The value of $34 \left| \frac{5z - 12}{5iz + 16} \right|^2$ when $z$ has maximum positive principal argument.

For maximum argument, $z$ lies at the point of tangency of the line from the origin to the circle.

Let $O$ be the origin, $C(5,0)$ the center, and $P$ the tangency point. In right triangle $OPC$,

$$OP = \sqrt{OC^2 - CP^2} = \sqrt{25 - 9} = 4$$

If $\theta$ is the argument of $z$, then

$$\sin \theta = \frac{CP}{OC} = \frac{3}{5}, \qquad \cos \theta = \frac{OP}{OC} = \frac{4}{5}$$

Hence,

$$z = OP(\cos \theta + i \sin \theta) = 4\left(\frac{4}{5} + i\frac{3}{5}\right) = \frac{16 + 12i}{5}$$

So,

$$5z = 16 + 12i$$

Now substitute into the required expression:

$$E = 34 \left| \frac{(16 + 12i) - 12}{i(16 + 12i) + 16} \right|^2$$ $$E = 34 \left| \frac{4 + 12i}{16i - 12 + 16} \right|^2 = 34 \left| \frac{4 + 12i}{4 + 16i} \right|^2$$

Using $\left|\frac{a}{b}\right|^2 = \frac{|a|^2}{|b|^2}$,

$$E = 34 \frac{|4 + 12i|^2}{|4 + 16i|^2} = 34 \frac{16 + 144}{16 + 256} = 34 \frac{160}{272}$$

Since $272 = 8 \times 34$,

$$E = \frac{34 \times 160}{8 \times 34} = \frac{160}{8} = 20$$

Therefore, the value is $20$. The correct option is D.

Given: The locus $|z-5| \le 3$ is a closed disk centered on the positive real axis.

Find: The extreme-point value of the given modulus expression.

The point with maximum positive principal argument on this disk is the tangency point from the origin to the boundary circle $|z-5|=3$. This is because rotating the ray from the origin counterclockwise, the last point of contact with the disk occurs at tangency.

From the right triangle formed by the origin, the center, and the tangency point,

$$OC = 5, \qquad CP = 3, \qquad OP = 4$$

Thus the tangency point has polar form

$$z = 4(\cos\theta + i\sin\theta)$$

with

$$\cos\theta = \frac{4}{5}, \qquad \sin\theta = \frac{3}{5}$$

Therefore,

$$z = 4\left(\frac{4}{5} + i\frac{3}{5}\right) = \frac{16+12i}{5}$$

Substituting this value of $z$ into the expression gives the required result:

$$34 \left| \frac{5z - 12}{5iz + 16} \right|^2 = 34 \left| \frac{4+12i}{4+16i} \right|^2 = 20$$

Therefore, the value is $20$.