Question:

If $|z - 25i| \leq 15$, the maximum $\arg(z) -$ minimum $\arg(z)$ is equal to

Updated On: Apr 15, 2025
  • 2cos-1 (3/5)
  • 2cos-1 (4/5)
  • π/2+cos-1 (3/5)
  • sin-1 (3/5) - cos-1 (3/5)
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The Correct Option is B

Solution and Explanation

We are given that \( |z - 25i| \leq 15 \), where \( z \) is a complex number. This describes a circle in the complex plane centered at \( 25i \) (which corresponds to the point \( (0, 25) \) in the complex plane) with radius 15.

Step 1: Geometrical interpretation
The inequality \( |z - 25i| \leq 15 \) represents the set of all complex numbers \( z \) whose distance from the point \( 25i \) is less than or equal to 15. This is a circle centered at \( (0, 25) \) with radius 15. 

Step 2: Maximum and minimum arguments of \( z \)
To find the maximum and minimum arguments of \( z \), we consider the lines from the center of the circle (i.e., the point \( 25i \)) to the points on the circle that are farthest and closest to the positive real axis. - The maximum argument corresponds to the direction that is furthest counterclockwise from the positive real axis. - The minimum argument corresponds to the direction that is furthest clockwise from the positive real axis. 

Step 3: Calculation of the angle
The maximum angular separation between the maximum and minimum arguments occurs along the vertical line through the center of the circle (since the center is on the imaginary axis). The distance from the origin to the center of the circle is \( 25 \) (the imaginary part of \( 25i \)), and the radius of the circle is \( 15 \). We can use the following relationship to calculate the angle between the maximum and minimum arguments: \[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{25}{\sqrt{25^2 + 15^2}} = \frac{25}{\sqrt{625 + 225}} = \frac{25}{\sqrt{850}} = \frac{25}{\sqrt{850}} = \frac{4}{5} \] Thus, the maximum difference between the arguments is: \[ \text{Angle} = 2 \cos^{-1} \left( \frac{4}{5} \right) \]

Answer:

\[ \boxed{2 \cos^{-1} \left( \frac{4}{5} \right)} \]

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