Complex NumbersHard

Question

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

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

Options

A.16
B.12
C.26
D.20

Solution

$$|z - 5| \leq 3 $$For $arg(z)$ to be maximum, z lies at P .

$${z \equiv (4cos\theta,4sin\theta) }{\equiv \left( 4 \cdot \left( \frac{4}{5} \right),4\left( \frac{3}{5} \right) \right) = \left( \frac{16}{5},\frac{12}{5} \right) = \frac{16}{5} + \frac{12i}{5} }$$Now, $34\left| \frac{5z - 12}{5iz + 16} \right|^{2} = 34\left| \frac{(16 + 12i) - 12}{(16i - 12) + 16} \right|^{2}$

$${= 34\left| \frac{4 + 12i}{16i + 4} \right|^{2} }{= 34\left( \frac{16 + 144}{256 + 16} \right) = 34\left( \frac{160}{272} \right) = 20}$$

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