Let z be the complex number satisfying $$|z-5|\leq 3$$ and having maximum positive principal argument. Then $$34|\frac{5z-12}{5iz+16}|^2$$ is equal to :

The region $$|z - 5| \leq 3$$ is the disk centered at $$(5, 0)$$ with radius 3, and the desired point is where the line from the origin is tangent to the circle.

Let the maximum argument be $$\alpha$$; since the distance from the origin to the center is 5 and the radius is 3, it follows that $$\sin\alpha = 3/5$$.

 $$\cos\alpha = 4/5$$.

The point of tangency $$z$$ lies where $$\vec{Oz}$$ makes angle $$\alpha$$ with the x-axis on the circle $$|z-5|=3$$.

The length of the tangent from the origin is $$\sqrt{25 - 9} = 4$$.

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

We then compute $$5z - 12 = 16 + 12i - 12 = 4 + 12i$$.

 $$5iz + 16 = 5i\left(\frac{16}{5} + \frac{12i}{5}\right) + 16 = 16i + 12i^2 + 16 = 16i - 12 + 16 = 4 + 16i$$.

 $$\left|\frac{4+12i}{4+16i}\right|^2 = \frac{16 + 144}{16 + 256} = \frac{160}{272} = \frac{10}{17}$$.

Thus, $$34 \times \frac{10}{17} = 20$$.