Question

Let \( r \) and \( \theta \) respectively be the modulus and amplitude of the complex number \( z = 2 - i \left( 2 \tan \frac{5\pi}{8} \right) \), then \( (r, \theta) \) is equal to

• A. \( \left( 2 \sec \frac{3\pi}{8}, \frac{3\pi}{8} \right) \)
• B. \( \left( 2 \sec \frac{3\pi}{8}, \frac{5\pi}{8} \right) \)
• C. \( \left( 2 \sec \frac{5\pi}{8}, \frac{3\pi}{8} \right) \)
• D. \( \left( 2 \sec \frac{11\pi}{8}, \frac{11\pi}{8} \right) \)

Answer 1

The Correct Option is A

To determine the modulus and amplitude of the complex number \( z = 2 - i \left( 2 \tan \frac{5\pi}{8} \right) \), we proceed as follows:

The given complex number is in the form \( z = a + bi \), where \( a = 2 \) and \( b = -2 \tan \frac{5\pi}{8} \).

To find the modulus \( r \) of \( z \), use the formula:

\(r = \sqrt{a^2 + b^2}\)

Substitute \( a \) and \( b \):

\(r = \sqrt{2^2 + \left(-2 \tan \frac{5\pi}{8}\right)^2}\)

\(r = \sqrt{4 + 4 \tan^2 \frac{5\pi}{8}}\)

\(r = 2 \sqrt{1 + \tan^2 \frac{5\pi}{8}}\)

Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \):

\(r = 2 \sec \frac{5\pi}{8}\)

Next, calculate the amplitude \(\theta\) using the formula \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\):

\(\theta = \tan^{-1}\left(\frac{-2 \tan \frac{5\pi}{8}}{2}\right) = \tan^{-1}\left(-\tan \frac{5\pi}{8}\right)\)

\(\theta = \pi - \frac{5\pi}{8} = \frac{3\pi}{8}\) because the complex number is in the second quadrant.

Thus, the pair \((r, \theta)\) is \(\left(2 \sec \frac{3\pi}{8}, \frac{3\pi}{8}\right)\).

The correct answer is therefore \(\left( 2 \sec \frac{3\pi}{8}, \frac{3\pi}{8} \right)\).

Answer 2

Let \( z = 2 - i \left( 2 \tan \frac{5\pi}{8} \right) = x + iy \).

Step 1. Calculating \( r \), the modulus of \( z \):**  
 \(r = \sqrt{x^2 + y^2}\)
 \(r = \sqrt{(2)^2 + \left( 2 \tan \frac{5\pi}{8} \right)^2}\)
 \(= 2 \sec \frac{5\pi}{8} = 2 \sec \left( \pi - \frac{3\pi}{8} \right) = 2 \sec \frac{3\pi}{8}\)
 

Step 2. Calculating \( \theta \), the amplitude of \( z \):
  \(\theta = \tan^{-1} \left( \frac{-2 \tan \frac{5\pi}{8}}{2} \right)\)
  \(= \tan^{-1} \left( -\tan \frac{5\pi}{8} \right) = \frac{3\pi}{8}\)
Therefore, \( (r, \theta) = \left( 2 \sec \frac{3\pi}{8}, \frac{3\pi}{8} \right) \).

The Correct Answer is:\(\left( 2 \sec \frac{3\pi}{8}, \frac{3\pi}{8} \right)\).