Question

Let \( w_1 \) be the point obtained by the rotation of \( z_1 = 5 + 4i \) about the origin through a right angle in the anticlockwise direction, and \( w_2 \) be the point obtained by the rotation of \( z_2 = 3 + 5i \) about the origin through a right angle in the clockwise direction. Then the principal argument of \( w_1 - w_2 \) is equal to:

• A. \( \pi - \tan^{-1}\left( \frac{8}{9} \right) \)
• B. \( \pi - \tan^{-1}\left( \frac{48}{9} \right) \)
• C. \( \pi - \tan^{-1}\left( \frac{33}{5} \right) \)
• D. \( \pi - \tan^{-1}\left( \frac{33}{5} \right) \)

Hint:

The rotation of a complex number by \( 90^{\circ} \) can be performed using multiplication by \( i \) (anticlockwise) or \( -i \) (clockwise). Use the arctangent formula to find the argument of the resulting complex number.

Answer 1

The Correct Option is A

Let \( w_1 \) and \( w_2 \) be the points obtained by the rotations of the complex numbers \( z_1 = 5 + 4i \) and \( z_2 = 3 + 5i \), respectively.
We are asked to find the principal argument of \( w_1 - w_2 \).
For \( w_1 \), the rotation is anticlockwise by \( 90^{\circ} \). The rotation of a complex number \( z = x + yi \) by \( 90^{\circ} \) anticlockwise is given by the transformation:
\[ w_1 = i \cdot z_1 = i \cdot (5 + 4i) = -4 + 5i \] For \( w_2 \), the rotation is clockwise by \( 90^{\circ} \). The rotation of a complex number \( z = x + yi \) by \( 90^{\circ} \) clockwise is given by the transformation: \[ w_2 = -i \cdot z_2 = -i \cdot (3 + 5i) = 5 + 3i \] Now, we need to compute the difference \( w_1 - w_2 \): \[ w_1 - w_2 = (-4 + 5i) - (5 + 3i) = -4 - 5 + (5 - 3)i = -9 + 2i \] The principal argument \( \theta \) of a complex number \( z = x + yi \) is given by: \[ \theta = \tan^{-1}\left( \frac{y}{x} \right) \] Thus, for \( w_1 - w_2 = -9 + 2i \): \[ \theta = \tan^{-1}\left( \frac{2}{-9} \right) \] Since \( w_1 - w_2 \) lies in the second quadrant, the principal argument is: \[ \text{Principal Argument} = \pi + \tan^{-1}\left( \frac{2}{9} \right) \] Thus, the correct option is \( \pi - \tan^{-1}\left( \frac{8}{9} \right) \).