A straight line is passing through the points represented by the complex numbers `a+ib` and `( 1)/( -a+ib)` , where `( a, b ) cancel(=)(0,0)`
Which one of the following is correct?

Options

A. It passes through the origin
B. It is parallel to the a=axis
C. It is parallel to the y-axis
D. It passes through (0,b)

Answer

To solve the problem, we need to find the equation of the straight line passing through two points represented by the complex numbers \( a + ib \) and \( \frac{1}{-a + ib} \).

### Step-by-Step Solution:

1. **Identify the Points**:
The first point is \( P_1 = a + ib \) which corresponds to the coordinates \( (a, b) \).
The second point is \( P_2 = \frac{1}{-a + ib} \).

2. **Rationalize the Second Point**:
To express \( P_2 \) in the form \( x + iy \), we need to rationalize the denominator:
\[ P_2 = \frac{1}{-a + ib} \cdot \frac{-a - ib}{-a - ib} = \frac{-a - ib}{(-a)^2 + (b)^2} = \frac{-a - ib}{a^2 + b^2} \]
This gives us:
\[ P_2 = \left(-\frac{a}{a^2 + b^2}\right) + i\left(-\frac{b}{a^2 + b^2}\right) \]
Hence, the coordinates of \( P_2 \) are:
\[ \left(-\frac{a}{a^2 + b^2}, -\frac{b}{a^2 + b^2}\right) \]

3. **Set Up the Equation of the Line**:
We have two points \( P_1(a, b) \) and \( P_2\left(-\frac{a}{a^2 + b^2}, -\frac{b}{a^2 + b^2}\right) \).
The slope \( m \) of the line passing through these points is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-\frac{b}{a^2 + b^2} - b}{-\frac{a}{a^2 + b^2} - a} \]

4. **Simplify the Slope**:
Let's simplify the numerator and denominator:
- **Numerator**:
\[ -\frac{b}{a^2 + b^2} - b = -\frac{b + b(a^2 + b^2)}{a^2 + b^2} = -\frac{b(1 + a^2 + b^2)}{a^2 + b^2} \]
- **Denominator**:
\[ -\frac{a}{a^2 + b^2} - a = -\frac{a + a(a^2 + b^2)}{a^2 + b^2} = -\frac{a(1 + a^2 + b^2)}{a^2 + b^2} \]

5. **Combine to Find the Slope**:
Thus, the slope \( m \) becomes:
\[ m = \frac{-\frac{b(1 + a^2 + b^2)}{a^2 + b^2}}{-\frac{a(1 + a^2 + b^2)}{a^2 + b^2}} = \frac{b}{a} \]

6. **Equation of the Line**:
Now, using the point-slope form of the line equation:
\[ y - b = \frac{b}{a}(x - a) \]
Rearranging gives:
\[ ay = bx \]
This indicates that the line passes through the origin (0, 0).

### Conclusion:
The straight line passes through the origin.