If z is purely imaginary and `Im (z) lt 0`, then `arg(i bar(z)) + arg(z)` is equal to

Options

A. `pi`
B. 0
C. `pi//2`
D. `-pi//2`

Answer

To solve the problem, we need to find the value of \( \arg(i \bar{z}) + \arg(z) \) given that \( z \) is purely imaginary and \( \text{Im}(z) < 0 \).

### Step-by-Step Solution:

1. **Express \( z \)**:
Since \( z \) is purely imaginary, we can write it as:
\[ z = 0 + ai \quad \text{where } a < 0 \]
This means \( z = ai \) where \( a \) is a negative real number.

**Hint**: Remember that a purely imaginary number has no real part.

2. **Find \( \bar{z} \)**:
The conjugate of \( z \) is:
\[ \bar{z} = -ai \]

**Hint**: The conjugate of a complex number flips the sign of the imaginary part.

3. **Calculate \( i \bar{z} \)**:
Now, we compute \( i \bar{z} \):
\[ i \bar{z} = i(-ai) = -a(i^2) = -a(-1) = a \]
Since \( a < 0 \), we have \( i \bar{z} = a \) which is a negative real number.

**Hint**: Multiplying by \( i \) rotates the complex number by \( 90^\circ \) counterclockwise.

4. **Find \( \arg(i \bar{z}) \)**:
The argument of a negative real number is:
\[ \arg(i \bar{z}) = \arg(a) = \pi \]

**Hint**: The argument of a negative real number is \( \pi \) radians.

5. **Find \( \arg(z) \)**:
Since \( z = ai \) and \( a < 0 \), the argument of \( z \) is:
\[ \arg(z) = \arg(ai) = -\frac{\pi}{2} \]

**Hint**: The argument of a purely imaginary number in the negative direction is \( -\frac{\pi}{2} \).

6. **Combine the arguments**:
Now we can combine the arguments:
\[ \arg(i \bar{z}) + \arg(z) = \pi + \left(-\frac{\pi}{2}\right) = \pi - \frac{\pi}{2} = \frac{\pi}{2} \]

**Hint**: When adding angles, ensure you keep track of the signs.

### Final Answer:
\[ \arg(i \bar{z}) + \arg(z) = \frac{\pi}{2} \]