Aspire Faculty ID #17928 · Topic: JEE Main 2026 (21 January Evening Shift) · Just now
JEE Main 2026 (21 January Evening Shift)

Let $z$ be the complex number satisfying $|z - 5| \le 3$ and having maximum positive principal argument. Then

$34\left|\frac{5z - 12}{5iz + 16}\right|^2$ is equal to:

Solution



$|z - 5| \le 3$

For $\arg(z)$ to be maximum, $z$ lies at point $P$

$z = (4, 4)$

$= \left(4\left(\frac{4}{5}\right),; 4\left(\frac{3}{5}\right)\right)$

$= \left(\frac{16}{5},; \frac{12}{5}\right)$

Now

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

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

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

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

$= 34\left(\frac{160}{272}\right) = 20$
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