Let z be the complex number satisfying |z-5| | vedclass

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(D) The condition $|z-5| \le 3$ represents a disk centered at $(5, 0)$ with radius $3$. For the argument of $z$ to be maximum, the line from the origi

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$20$

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(D) The condition $|z-5| \le 3$ represents a disk centered at $(5, 0)$ with radius $3$. For the argument of $z$ to be maximum, the line from the origin must be tangent to the circle at point $P(z)$.In the right-angled triangle formed by the origin $(0, 0)$, the center $(5, 0)$, and the point $P(z)$, the hypotenuse is $5$ and the radius is $3$. Thus, the distance from the origin to $P$ is $\sqrt{5^2 - 3^2} = 4$.Let $z = x + iy$. From the geometry, $\cos 	heta = \frac{4}{5}$ and $\sin 	heta = \frac{3}{5}$.Thus, $z = 4(\cos 	heta + i \sin 	heta) = 4(\frac{4}{5} + i \frac{3}{5}) = \frac{16}{5} + i \frac{12}{5}$.Now, substitute $z$ into the expression:$5z - 12 = 5(\frac{16}{5} + i \frac{12}{5}) - 12 = 16 + 12i - 12 = 4 + 12i$.$5iz + 16 = 5i(\frac{16}{5} + i \frac{12}{5}) + 16 = 16i - 12 + 16 = 4 + 16i$.Then, $|\frac{5z-12}{5iz+16}|^2 = |\frac{4+12i}{4+16i}|^2 = |\frac{1+3i}{1+4i}|^2 = \frac{1^2 + 3^2}{1^2 + 4^2} = \frac{1+9}{1+16} = \frac{10}{17}$.Finally, $34 	imes \frac{10}{17} = 2 	imes 10 = 20$.

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

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