Let z be a purely imaginary number such that `"lm"(z) gt 0`. Then, arg (z) is equal to

Options

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

Explanation

To solve the problem, we need to find the argument of a purely imaginary number \( z \) where the imaginary part is greater than zero. Let's break this down step by step.

### Step-by-Step Solution:

1. **Understanding Purely Imaginary Numbers**:
A purely imaginary number can be represented as:
\[ z = 0 + ib \]
where \( b \) is a real number.

2. **Condition Given**:
We are given that the imaginary part of \( z \) is greater than zero:
\[ \text{Im}(z) > 0 \implies b > 0 \]
Therefore, we can express \( z \) as:
\[ z = ib \quad \text{(where \( b > 0 \))} \]

3. **Identifying the Position in the Complex Plane**:
In the complex plane, the real part (which is 0) is on the x-axis, and the imaginary part (which is \( b \)) is on the y-axis. Since \( b > 0 \), the point \( z \) lies on the positive y-axis.

4. **Finding the Argument**:
The argument of a complex number \( z = x + iy \) is given by:
\[ \text{arg}(z) = \tan^{-1}\left(\frac{y}{x}\right) \]
In our case, since \( x = 0 \) and \( y = b \) (where \( b > 0 \)), we have:
\[ \text{arg}(z) = \tan^{-1}\left(\frac{b}{0}\right) \]
The tangent of an angle approaches infinity as we approach \( 90^\circ \) or \( \frac{\pi}{2} \) radians.

5. **Conclusion**:
Therefore, the argument of \( z \) is:
\[ \text{arg}(z) = \frac{\pi}{2} \]

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