If arg`(z-a)/(z+a)=+-pi/2`, where a is a fixed real number, then the locus of z is

Options

A. a staight line
B. a circle with center at the origin and radius a
C. a circle with center on y-axis
D. none of these

Explanation

To solve the problem, we need to analyze the given condition involving the argument of a complex number. The condition states:

\[ \arg\left(\frac{z - a}{z + a}\right) = \pm \frac{\pi}{2} \]

where \( a \) is a fixed real number.

### Step-by-Step Solution:

1. **Understanding the Argument Condition**:
The condition \(\arg\left(\frac{z - a}{z + a}\right) = \pm \frac{\pi}{2}\) implies that the complex number \(\frac{z - a}{z + a}\) lies on the imaginary axis. This is because an argument of \(\frac{\pi}{2}\) corresponds to a point on the positive imaginary axis, and an argument of \(-\frac{\pi}{2}\) corresponds to a point on the negative imaginary axis.

2. **Setting Up the Equation**:
We can rewrite the condition as:
\[ \frac{z - a}{z + a} = iy \quad \text{for some real } y \]
This leads to two cases:
- Case 1: \(\frac{z - a}{z + a} = i\)
- Case 2: \(\frac{z - a}{z + a} = -i\)

3. **Solving Case 1**:
For the first case, we have:
\[ z - a = i(z + a) \]
Rearranging gives:
\[ z - iz = a + ia \]
\[ z(1 - i) = a(1 + i) \]
Thus,
\[ z = \frac{a(1 + i)}{1 - i} \]
To simplify, multiply the numerator and denominator by the conjugate of the denominator:
\[ z = \frac{a(1 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{a(1 + 2i - 1)}{1 + 1} = \frac{a(2i)}{2} = ai \]

4. **Solving Case 2**:
For the second case, we have:
\[ z - a = -i(z + a) \]
Rearranging gives:
\[ z + iz = -a + ia \]
\[ z(1 + i) = -a(1 - i) \]
Thus,
\[ z = \frac{-a(1 - i)}{1 + i} \]
Again, simplifying by multiplying by the conjugate:
\[ z = \frac{-a(1 - i)(1 - i)}{(1 + i)(1 - i)} = \frac{-a(1 - 2i - 1)}{1 + 1} = \frac{-a(-2i)}{2} = -ai \]

5. **Finding the Locus**:
From both cases, we find that \( z \) can take the values \( ai \) and \( -ai \). This means that the locus of \( z \) is along the imaginary axis, specifically the points where the real part is zero and the imaginary part is \( \pm a \).

### Conclusion:
The locus of \( z \) is a vertical line segment on the imaginary axis between the points \( ai \) and \( -ai \).