If `z = ( -2 ( 1+ 2i))/( ( 3+i))`, where `i = sqrt( -1)`, then the argument `theta ( - pi lt theta le pi )` of z is

Options

A. A) `( 3pi )/( 4)`
B. B) `( pi )/( 4)`
C. C) `( 5pi )/( 6)`
D. D) `- (3pi)/( 4)`

Answer

To find the argument \( \theta \) of the complex number \( z = \frac{-2(1 + 2i)}{3 + i} \), we will follow these steps:

### Step 1: Simplify the expression for \( z \)
We start with the expression:
\[ z = \frac{-2(1 + 2i)}{3 + i} \]
To simplify this, we multiply the numerator and the denominator by the conjugate of the denominator, which is \( 3 - i \):
\[ z = \frac{-2(1 + 2i)(3 - i)}{(3 + i)(3 - i)} \]

### Step 2: Calculate the denominator
The denominator is:
\[ (3 + i)(3 - i) = 3^2 - i^2 = 9 - (-1) = 9 + 1 = 10 \]

### Step 3: Calculate the numerator
Now, we calculate the numerator:
\[ -2(1 + 2i)(3 - i) = -2[(1)(3) + (1)(-i) + (2i)(3) + (2i)(-i)] \]
Calculating each term:
- \( 1 \cdot 3 = 3 \)
- \( 1 \cdot -i = -i \)
- \( 2i \cdot 3 = 6i \)
- \( 2i \cdot -i = -2i^2 = 2 \) (since \( i^2 = -1 \))

Combining these:
\[ 3 - i + 6i + 2 = 5 + 5i \]
Thus, the numerator becomes:
\[ -2(5 + 5i) = -10 - 10i \]

### Step 4: Combine the results
Now we can write \( z \):
\[ z = \frac{-10 - 10i}{10} = -1 - i \]

### Step 5: Find the argument \( \theta \)
The argument \( \theta \) of a complex number \( z = x + yi \) is given by:
\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]
Here, \( x = -1 \) and \( y = -1 \):
\[ \theta = \tan^{-1}\left(\frac{-1}{-1}\right) = \tan^{-1}(1) \]
The angle whose tangent is 1 is \( \frac{\pi}{4} \), but since both \( x \) and \( y \) are negative, \( z \) is in the third quadrant. Therefore, we add \( \pi \) to \( \frac{\pi}{4} \):
\[ \theta = \frac{\pi}{4} + \pi = \frac{5\pi}{4} \]

### Final Answer
Thus, the argument \( \theta \) of \( z \) is:
\[ \theta = \frac{5\pi}{4} \]