Question

If \( \sqrt{5} - i\sqrt{15} = r(\cos\theta + i\sin\theta), -\pi < \theta < \pi, \) then

\[ r^2(\sec\theta + 3\csc^2\theta) = \]

• A. 40
• B. 60
• C. 120
• D. 180

Hint:

For complex numbers, ensure the trigonometric functions are evaluated correctly and watch for signs during angle calculations.

Answer 1

The Correct Option is C

To solve the problem, we need to express the complex number \( \sqrt{5} - i\sqrt{15} \) in polar form and then compute \( r^2(\sec \theta + 3\csc^2 \theta) \). Step 1: Express \( \sqrt{5} - i\sqrt{15} \) in polar form The polar form of a complex number \( z = a + ib \) is given by: \[ z = r(\cos \theta + i\sin \theta), \] where:
\( r = \sqrt{a^2 + b^2} \) is the modulus,
\( \theta = \tan^{-1}\left(\frac{b}{a}\right) \) is the argument.
For \( z = \sqrt{5} - i\sqrt{15} \):
\( a = \sqrt{5} \),
\( b = -\sqrt{15} \).
Compute \( r \): \[ r = \sqrt{a^2 + b^2} = \sqrt{(\sqrt{5})^2 + (-\sqrt{15})^2} = \sqrt{5 + 15} = \sqrt{20} = 2\sqrt{5}. \] Compute \( \theta \): \[ \theta = \tan^{-1}\left(\frac{b}{a}\right) = \tan^{-1}\left(\frac{-\sqrt{15}}{\sqrt{5}}\right) = \tan^{-1}(-\sqrt{3}). \] Since \( \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \), and \( -\pi<\theta<\pi \), we have: \[ \theta = -\frac{\pi}{3}. \] Thus, the polar form of \( z \) is: \[ z = 2\sqrt{5}\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right). \] Step 2: Compute \( r^2(\sec \theta + 3\csc^2 \theta) \) We are tasked with finding: \[ r^2(\sec \theta + 3\csc^2 \theta). \] Compute \( r^2 \): \[ r^2 = (2\sqrt{5})^2 = 4 \times 5 = 20. \] Compute \( \sec \theta \): \[ \sec \theta = \frac{1}{\cos \theta}. \] At \( \theta = -\frac{\pi}{3} \), \( \cos\left(-\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \). Thus: \[ \sec \theta = \frac{1}{\frac{1}{2}} = 2. \] Compute \( \csc^2 \theta \): \[ \csc \theta = \frac{1}{\sin \theta}. \] At \( \theta = -\frac{\pi}{3} \), \( \sin\left(-\frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \). Thus: \[ \csc \theta = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}}. \] Squaring this: \[ \csc^2 \theta = \left(-\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}. \] Combine the terms: \[ \sec \theta + 3\csc^2 \theta = 2 + 3 \times \frac{4}{3} = 2 + 4 = 6. \] Multiply by \( r^2 \): \[ r^2(\sec \theta + 3\csc^2 \theta) = 20 \times 6 = 120. \] Final Answer: \[ \boxed{120} \]