Question

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 :

• A. 12
• B. 16
• C. 26
• D. 20

Hint:

The point of maximum argument for a circle $|z-z_0|=r$ (where $z_0$ is real) satisfies $\sin \theta = r/|z_0|$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The condition $|z - 5| \le 3$ represents a disk of radius 3 centered at $(5, 0)$ in the complex plane. The maximum argument occurs when $z$ lies at the point of tangency of a line from the origin to the circle.

Step 2: Detailed Explanation:
Let $O$ be the origin, $C(5, 0)$ be the center, and $P$ be the point of tangency.
In right triangle $\triangle OPC$: $OC = 5$, $CP = 3$.
$OP = \sqrt{OC^2 - CP^2} = \sqrt{25 - 9} = 4$.
If $\theta$ is the argument of $z$, then $\sin \theta = \frac{CP}{OC} = \frac{3}{5}$ and $\cos \theta = \frac{OP}{OC} = \frac{4}{5}$.
So, $z = OP(\cos \theta + i \sin \theta) = 4 \left( \frac{4}{5} + i \frac{3}{5} \right) = \frac{16 + 12i}{5}$.
Substitute $5z = 16 + 12i$ into the expression:
\[ E = 34 \left| \frac{(16 + 12i) - 12}{i(16 + 12i) + 16} \right|^2 \]
\[ E = 34 \left| \frac{4 + 12i}{16i - 12 + 16} \right|^2 = 34 \left| \frac{4 + 12i}{4 + 16i} \right|^2 \]
\[ E = 34 \frac{|4 + 12i|^2}{|4 + 16i|^2} = 34 \frac{16 + 144}{16 + 256} = 34 \frac{160}{272} \]
Since $272 = 8 \times 34$:
\[ E = \frac{34 \times 160}{8 \times 34} = \frac{160}{8} = 20. \]

Step 3: Final Answer:
The value is 20.