If `r=abs(z),theta="arg"(z)"and"z=2-2itan((5pi)/8)` then find `(r, theta)`

Options

A. `(2sec((5pi)/(8),(3pi)/8)`
B. `(2sec((3pi)/(8),(3pi)/8)`
C. `(2tan((3pi)/(8),(5pi)/8)`
D. `(2tan((3pi)/(8),(3pi)/8)`

Answer

To solve the problem, we need to find the values of \( r \) and \( \theta \) given that \( z = 2 - 2i \tan\left(\frac{5\pi}{8}\right) \).

### Step-by-Step Solution:

1. **Identify the given complex number**:
\[ z = 2 - 2i \tan\left(\frac{5\pi}{8}\right) \]

2. **Express \( z \) in the form \( a + bi \)**:
Here, \( a = 2 \) and \( b = -2 \tan\left(\frac{5\pi}{8}\right) \).

3. **Calculate the modulus \( r \)**:
The modulus \( r \) of a complex number \( z = a + bi \) is given by:
\[ r = |z| = \sqrt{a^2 + b^2} \]
Substitute \( a \) and \( b \):
\[ r = \sqrt{2^2 + \left(-2 \tan\left(\frac{5\pi}{8}\right)\right)^2} \]
\[ r = \sqrt{4 + 4 \tan^2\left(\frac{5\pi}{8}\right)} \]
\[ r = 2\sqrt{1 + \tan^2\left(\frac{5\pi}{8}\right)} \]
Using the identity \( 1 + \tan^2(\theta) = \sec^2(\theta) \):
\[ r = 2 \sec\left(\frac{5\pi}{8}\right) \]

4. **Calculate the argument \( \theta \)**:
The argument \( \theta \) of a complex number \( z = a + bi \) is given by:
\[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \]
Substitute \( a \) and \( b \):
\[ \theta = \tan^{-1}\left(\frac{-2 \tan\left(\frac{5\pi}{8}\right)}{2}\right) \]
\[ \theta = \tan^{-1}\left(-\tan\left(\frac{5\pi}{8}\right)\right) \]
Since \( \tan\left(\frac{5\pi}{8}\right) \) is positive, \( -\tan\left(\frac{5\pi}{8}\right) \) is negative, which means \( \theta \) will be in the fourth quadrant:
\[ \theta = -\frac{5\pi}{8} \]

5. **Final result**:
Thus, the values of \( r \) and \( \theta \) are:
\[ (r, \theta) = \left(2 \sec\left(\frac{5\pi}{8}\right), -\frac{5\pi}{8}\right) \]