Complex Numbers Part 2

Concept:The region $|z-5| \le 3$ is a closed disc of radius 3 centered at $(5,0)$ on the complex plane. The point in this disc with the maximum positive principal argument lies on the boundary and is tangent to the disc from the origin.Explanation:For maximum argument, the point $z$ is on the circle $|z-5|=3$ and the line from origin to $z$ is tangent to the circle. The distance from origin to center is $5$, radius is $3$, so $\sin\theta = \frac{3}{5}$, $\cos\theta = \frac{4}{5}$. Thus $z = |z| (\cos\theta + i\sin\theta) = 4\left(\frac{4}{5} + i\frac{3}{5}\right) = \frac{16}{5} + \frac{12i}{5}$. Hence $5z = 16 + 12i$. Now evaluate the expression: $\frac{5z-12}{5iz+16} = \frac{(16+12i)-12}{i(16+12i)+16} = \frac{4+12i}{16i-12+16} = \frac{4+12i}{4+16i} = \frac{1+3i}{1+4i}$. Its magnitude squared: $\left|\frac{1+3i}{1+4i}\right|^2 = \frac{1^2+3^2}{1^2+4^2} = \frac{10}{17}$. Therefore $34 \left| \frac{5z-12}{5iz+16} \right|^2 = 34 \times \frac{10}{17} = 20$. Answer:$20$ (Option A)