Locus of Complex Numbers in Given Parameterized Set

To determine the locus of the complex number $z = x + iy \in S$, we start by expressing $z$ in terms of its real and imaginary parts:

$z = \frac{1}{a + ibt}$

Multiplying the numerator and denominator by the complex conjugate $a - ibt$:

$z = \frac{a - ibt}{(a + ibt)(a - ibt)} = \frac{a - ibt}{a^2 + b^2 t^2}$

Since $z = x + iy$, equating the real and imaginary parts gives:

$x = \frac{a}{a^2 + b^2 t^2}$ $y = \frac{-bt}{a^2 + b^2 t^2}$

Now, we analyze the locus under different conditions for $a$ and $b$:

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Case 1: $a \neq 0$ and $b \neq 0$

Calculate $x^2 + y^2$:

$x^2 + y^2 = \left( \frac{a}{a^2 + b^2 t^2} \right)^2 + \left( \frac{-bt}{a^2 + b^2 t^2} \right)^2 = \frac{a^2 + b^2 t^2}{(a^2 + b^2 t^2)^2} = \frac{1}{a^2 + b^2 t^2}$

Since $x = \frac{a}{a^2 + b^2 t^2}$, we can substitute $\frac{1}{a^2 + b^2 t^2} = \frac{x}{a}$:

$x^2 + y^2 = \frac{x}{a}$ $x^2 - \frac{x}{a} + y^2 = 0$

Completing the square for $x$:

$\left( x - \frac{1}{2a} \right)^2 + y^2 = \frac{1}{4a^2}$

This is the equation of a circle with:

• Centre: $\left( \frac{1}{2a}, 0 \right)$
• Radius: $R = \sqrt{\frac{1}{4a^2}} = \frac{1}{2|a|}$

1. For $a > 0, b \neq 0$:

   • Centre $= \left( \frac{1}{2a}, 0 \right)$
   • Radius $= \frac{1}{2a}$

   Thus, Option (A) is correct.

2. For $a < 0, b \neq 0$:

   • Centre $= \left( \frac{1}{2a}, 0 \right)$
   • Radius $= -\frac{1}{2a}$

   Option (B) states the centre as $\left( -\frac{1}{2a}, 0 \right)$, which is incorrect since $\frac{1}{2a} \neq -\frac{1}{2a}$ for $a < 0$. Thus, Option (B) is incorrect.

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Case 2: $a \neq 0$ and $b = 0$

Substituting $b = 0$ into the equations for $x$ and $y$:

$x = \frac{a}{a^2} = \frac{1}{a}$ $y = 0$

Since $y = 0$, the point $(x, y) = \left( \frac{1}{a}, 0 \right)$ lies on the $x$-axis for any non-zero real parameter $t$.

Thus, Option (C) is correct.

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Case 3: $a = 0$ and $b \neq 0$

Substituting $a = 0$ into the equations for $x$ and $y$:

$x = 0$ $y = -\frac{1}{bt}$

Since $t \in \mathbb{R} \setminus \{0\}$ and $b \neq 0$, $y$ takes all non-zero real values while $x = 0$. Therefore, $(x, y)$ lies on the $y$-axis.

Thus, Option (D) is correct.

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Conclusion

The correct options are A, C, and D.