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 :
Region for Complex Number z
The condition $|z - 5| \le 3$ defines a closed disk in the complex plane. This disk is centered at $5$ (corresponding to the point $(5, 0)$) and has a radius of $3$. All complex numbers $z$ satisfying this condition lie within or on the boundary of this circle.
Finding z with Maximum Positive Argument
We seek the complex number $z$ within the disk $|z - 5| \le 3$ that possesses the maximum positive principal argument. The principal argument $\theta$ satisfies $-\pi < \theta \le \pi$. Since the disk is centered at $(5, 0)$ with radius $3$, all points $z = x + iy$ in the disk have $x \ge 2$. Consequently, all points are located in the first or fourth quadrant, meaning their arguments lie within the range $(-\pi/2, \pi/2)$. The maximum positive argument occurs at a point on the boundary circle, $(x-5)^2 + y^2 = 9$, where the line segment connecting the origin to $z$ is tangent to the circle.
Let $z = x + iy$. For tangency, the radius vector from the center $(5, 0)$ to $z = (x, y)$ must be perpendicular to the vector from the origin to $z = (x, y)$. The dot product of these vectors, $(x-5, y)$ and $(x, y)$, must be zero:
$x(x-5) + y^2 = 0$ $x^2 - 5x + y^2 = 0 \quad (*)$
Since $z$ is on the circle boundary, it also satisfies $(x-5)^2 + y^2 = 9$. Expanding this yields:
$x^2 - 10x + 25 + y^2 = 9 \quad (**)$
Substitute $y^2 = 5x - x^2$ from (*) into (**):
$x^2 - 10x + 25 + (5x - x^2) = 9$ $-5x + 25 = 9$ $-5x = -16$ $x = \frac{16}{5}$
Find $y$ using $y^2 = 5x - x^2$:
$y^2 = 5\left(\frac{16}{5}\right) - \left(\frac{16}{5}\right)^2 = 16 - \frac{256}{25} = \frac{400 - 256}{25} = \frac{144}{25}$ $y = \pm \frac{12}{5}$
To achieve the maximum positive argument, we select the positive value for $y$. Therefore, the complex number is:
$z = \frac{16}{5} + i \frac{12}{5}$
Calculating the Expression Value
The expression to evaluate is $34 \left| \frac{5z - 12}{5iz + 16} \right|^2$. Substitute the determined value of $z$.
Calculate the numerator $5z - 12$:
$5z - 12 = 5\left(\frac{16}{5} + i \frac{12}{5}\right) - 12 = (16 + 12i) - 12 = 4 + 12i$
Calculate the denominator $5iz + 16$:
$5iz + 16 = 5i\left(\frac{16}{5} + i \frac{12}{5}\right) + 16 = i(16 + 12i) + 16 = 16i + 12i^2 + 16$ $= 16i - 12 + 16 = 4 + 16i$
Compute the fraction:
$\frac{5z - 12}{5iz + 16} = \frac{4 + 12i}{4 + 16i} = \frac{4(1 + 3i)}{4(1 + 4i)} = \frac{1 + 3i}{1 + 4i}$
Determine the squared magnitude of this fraction:
$\left| \frac{1 + 3i}{1 + 4i} \right|^2 = \frac{|1 + 3i|^2}{|1 + 4i|^2} = \frac{1^2 + 3^2}{1^2 + 4^2} = \frac{1 + 9}{1 + 16} = \frac{10}{17}$
Finally, compute the value of the entire expression:
$34 \left| \frac{5z - 12}{5iz + 16} \right|^2 = 34 \times \frac{10}{17} = 2 \times 10 = 20$
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