Let `0 ne a, 0 ne b in R`. Suppose
`S={z in C, z=1/(a+ibt)t in R, t ne 0}`, where `i=sqrt(-1)`. If `z=x+iy` and `z in S`, then `(x,y)` lies on

Options

A. the circle with radius `1/(2a)` and centre `(1/(2a),0)` for `a gt 0, b ne 0`
B. the circle with radius `-1/(2a)` and centre`(-1/(2a),0)` for `a lt 0,b ne 0`
C. the X-axis for `a ne 0,b=0`
D. the Y-axis for `a=0, b ne 0`

Answer

To solve the problem, we start with the given set \( S = \{ z \in \mathbb{C} : z = \frac{1}{a + ibt}, t \in \mathbb{R}, t \neq 0 \} \).

1. **Express \( z \) in terms of \( t \)**:
We have:
\[ z = \frac{1}{a + ibt} \]
To simplify this expression, we multiply the numerator and the denominator by the conjugate of the denominator:
\[ z = \frac{1}{a + ibt} \cdot \frac{a - ibt}{a - ibt} = \frac{a - ibt}{a^2 + (bt)^2} \]

2. **Separate \( z \) into real and imaginary parts**:
From the above expression, we can separate \( z \) into its real and imaginary components:
\[ z = \frac{a}{a^2 + (bt)^2} - i \frac{bt}{a^2 + (bt)^2} \]
Let \( x = \frac{a}{a^2 + (bt)^2} \) and \( y = -\frac{bt}{a^2 + (bt)^2} \).

3. **Relate \( x \) and \( y \)**:
We can express \( y \) in terms of \( x \):
From \( x = \frac{a}{a^2 + (bt)^2} \), we can rearrange it to find \( (bt)^2 \):
\[ (bt)^2 = \frac{a}{x} - a^2 \]
Now substituting this into the equation for \( y \):
\[ y = -\frac{bt}{a^2 + (bt)^2} = -\frac{bt}{\frac{a}{x}} = -\frac{btx}{a} \]

4. **Eliminate \( t \)**:
To eliminate \( t \), we can express \( t \) in terms of \( x \) and \( y \):
From \( y = -\frac{bt}{a^2 + (bt)^2} \), we can rearrange to find \( t \):
\[ t = -\frac{ay}{b} \]
Substitute \( t \) back into the equation for \( x \):
\[ x = \frac{a}{a^2 + b^2 \left(-\frac{ay}{b}\right)^2} = \frac{a}{a^2 + \frac{a^2y^2}{b^2}} = \frac{a}{a^2(1 + \frac{y^2}{b^2})} = \frac{b^2}{b^2 + y^2} \]

5. **Identify the relationship**:
The relationship between \( x \) and \( y \) can be expressed as:
\[ x = \frac{b^2}{b^2 + y^2} \]
Rearranging gives:
\[ x(b^2 + y^2) = b^2 \implies b^2x + xy^2 = b^2 \]
This can be rearranged to:
\[ xy^2 + b^2x - b^2 = 0 \]
This is a quadratic equation in \( y \).

6. **Conclusion**:
The points \( (x, y) \) lie on a curve defined by the above equation, which describes a relationship between \( x \) and \( y \).