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

Options

A. `(2 sec frac{3pi}{8}, frac{3pi}{8})`
B. `(2 sec frac{3pi}{8}, frac{5pi}{8})`
C. `(2 sec frac{5pi}{8}, frac{3pi}{8})`
D. `(2 sec frac{11pi}{8}, frac{11pi}{8})`

Answer

To find the modulus \( r \) and amplitude \( \theta \) of the complex number \( z = 2 - i(2 \tan \frac{5\pi}{8}) \), we will follow these steps:

### Step 1: Identify the components of the complex number
The complex number can be expressed in the form \( z = x + iy \), where:
- \( x = 2 \)
- \( y = -2 \tan \frac{5\pi}{8} \)

### Step 2: Calculate \( y \)
First, we need to calculate \( y \):
\[ y = -2 \tan \frac{5\pi}{8} \]
Using the identity \( \tan \left( \frac{5\pi}{8} \right) = -\tan \left( \frac{3\pi}{8} \right) \):
\[ y = -2 \left(-\tan \frac{3\pi}{8}\right) = 2 \tan \frac{3\pi}{8} \]

### Step 3: Calculate the modulus \( r \)
The modulus \( r \) of the complex number is given by:
\[ r = \sqrt{x^2 + y^2} \]
Substituting the values of \( x \) and \( y \):
\[ r = \sqrt{2^2 + (2 \tan \frac{3\pi}{8})^2} \]
\[ = \sqrt{4 + 4 \tan^2 \frac{3\pi}{8}} \]
\[ = \sqrt{4(1 + \tan^2 \frac{3\pi}{8})} \]
Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \):
\[ = \sqrt{4 \sec^2 \frac{3\pi}{8}} = 2 \sec \frac{3\pi}{8} \]

### Step 4: Calculate the amplitude \( \theta \)
The amplitude \( \theta \) is given by:
\[ \theta = \tan^{-1} \left( \frac{y}{x} \right) \]
Substituting the values of \( y \) and \( x \):
\[ \theta = \tan^{-1} \left( \frac{2 \tan \frac{3\pi}{8}}{2} \right) = \tan^{-1} \left( \tan \frac{3\pi}{8} \right) \]
Since \( \tan^{-1} \tan \theta = \theta \) when \( \theta \) is in the range of \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \):
\[ \theta = \frac{3\pi}{8} \]

### Final Result
Thus, the modulus and amplitude of the complex number \( z \) are:
\[ (r, \theta) = \left( 2 \sec \frac{3\pi}{8}, \frac{3\pi}{8} \right) \]