JEE Main 2026 — 21 January, Evening Shift — Question 19

$|z-5| \leq 3$ For $\arg (\mathrm{z})$ to be maximum, z lies at P .

$$\begin{aligned} & z \equiv(4 \cos \theta, 4 \sin \theta) & \equiv\left(4 \cdot\left(\frac{4}{5}\right), 4\left(\frac{3}{5}\right)\right)=\left(\frac{16}{5}, \frac{12}{5}\right)=\frac{16}{5}+\frac{12 \mathrm{i}}{5} \end{aligned}$$

Now, $34\left|\frac{5 z-12}{5 i z+16}\right|^{2}=34\left|\frac{(16+12 i)-12}{(16 i-12)+16}\right|^{2}$ $=34\left|\frac{4+12 i}{16 i+4}\right|^{2}$

$=34\left(\frac{16+144}{256+16}\right)=34\left(\frac{160}{272}\right)=20$