If `z=(3-i)/(2+i)+(3+i)/(2-i)` then value of arg (zi) is

Options

A. 0
B. `pi/6`
C. `pi/3`
D. `pi/2`

Answer

To solve the problem \( z = \frac{3-i}{2+i} + \frac{3+i}{2-i} \) and find the value of \( \arg(zi) \), we will go through the following steps:

### Step 1: Simplify \( z \)

We start with the expression for \( z \):

\[ z = \frac{3-i}{2+i} + \frac{3+i}{2-i} \]

To simplify this, we will find a common denominator. The common denominator will be \( (2+i)(2-i) \).

### Step 2: Calculate the denominators

Calculating the denominator:

\[ (2+i)(2-i) = 2^2 - i^2 = 4 - (-1) = 4 + 1 = 5 \]

### Step 3: Rewrite \( z \) with the common denominator

Now we can rewrite \( z \):

\[ z = \frac{(3-i)(2-i) + (3+i)(2+i)}{5} \]

### Step 4: Expand the numerators

Now we will expand both numerators:

1. For \( (3-i)(2-i) \):
\[ = 6 - 3i - 2i + i^2 = 6 - 5i - 1 = 5 - 5i \]

2. For \( (3+i)(2+i) \):
\[ = 6 + 3i + 2i + i^2 = 6 + 5i - 1 = 5 + 5i \]

### Step 5: Combine the numerators

Now combine the results from the two expansions:

\[ z = \frac{(5 - 5i) + (5 + 5i)}{5} = \frac{10}{5} = 2 \]

### Step 6: Find \( zi \)

Now, we need to find \( zi \):

\[ zi = 2i \]

### Step 7: Find the argument of \( zi \)

The argument of a complex number \( a + bi \) is given by \( \tan^{-1}(\frac{b}{a}) \). Here, \( a = 0 \) and \( b = 2 \):

\[ \arg(zi) = \tan^{-1}\left(\frac{2}{0}\right) \]

Since the real part \( a = 0 \) and the imaginary part \( b > 0 \), the argument is:

\[ \arg(zi) = \frac{\pi}{2} \]

### Final Answer

Thus, the value of \( \arg(zi) \) is:

\[ \frac{\pi}{2} \]