Properties of the Solution Set for a Complex Equation

To determine which of the statements are true, we analyze the given complex equation: $sz + t\bar{z} + r = 0 \quad \text{--- (1)}$

Taking the complex conjugate of both sides of equation (1), we obtain: $\bar{s}\bar{z} + \bar{t}z + \bar{r} = 0 \implies \bar{t}z + \bar{s}\bar{z} + \bar{r} = 0 \quad \text{--- (2)}$

We can write equations (1) and (2) as a system of linear equations in $z$ and $\bar{z}$:

$$\begin{pmatrix} s & t \\ \bar{t} & \bar{s} \end{pmatrix} \begin{pmatrix} z \\ \bar{z} \end{pmatrix} = \begin{pmatrix} -r \\ -\bar{r} \end{pmatrix}$$

The determinant of the coefficient matrix is: $D = s\bar{s} - t\bar{t} = |s|^2 - |t|^2$

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Analysis of Options:

Option (A):

• If $|s| \neq |t|$, then $D \neq 0$. The system has a unique solution given by Cramer's rule: $z = \frac{-r\bar{s} + t\bar{r}}{|s|^2 - |t|^2} = \frac{t\bar{r} - r\bar{s}}{|s|^2 - |t|^2}$
• If $|s| = |t|$, then $D = 0$. A system of linear equations with $D = 0$ either has no solution or infinitely many solutions; it can never have a unique solution.
• Therefore, if $L$ contains precisely one element, we must have $|s| \neq |t|$.

Thus, Option (A) is TRUE.

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Option (B):

• Let $s = 1$, $t = 1$, and $r = i$. Note that $|s| = |t| = 1$.
• Substituting these into the original equation gives: $z + \bar{z} + i = 0$
• Writing $z = x + iy$ (where $x, y \in \mathbb{R}$), we have: $2x + i = 0 \implies 2x = -i$
• Since $x \in \mathbb{R}$, this equation has no solution, so $L = \emptyset$ (contains $0$ elements).
• Thus, $|s| = |t|$ does not guarantee infinitely many elements.

Thus, Option (B) is FALSE.

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Option (C):

• Writing $z = x + iy$, $s = s_1 + i s_2$, $t = t_1 + i t_2$, and $r = r_1 + i r_2$, equation (1) splits into two real linear equations in $x$ and $y$: $\text{Re}(sz + t\bar{z} + r) = (s_1 + t_1)x + (-s_2 + t_2)y + r_1 = 0$ $\text{Im}(sz + t\bar{z} + r) = (s_2 + t_2)x + (s_1 - t_1)y + r_2 = 0$
• Geometric locus of $L$ in the Cartesian plane $\mathbb{R}^2$ can only be:
  1. An empty set $\emptyset$,
  2. A single point $(x_0, y_0)$,
  3. A straight line.
• The set $\{z : |z - 1 + i| = 5\}$ represents a circle with center $(1, -1)$ and radius $5$.
• A circle can intersect a straight line at most $2$ times, a single point at most $1$ time, and an empty set $0$ times.
• Hence, $L \cap \{z : |z - 1 + i| = 5\}$ contains at most $2$ elements.

Thus, Option (C) is TRUE.

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Option (D):

• As shown in the geometric classification above, $L$ can only be an empty set, a single point, or a straight line.
• If $L$ has more than one element, it cannot be empty or a single point; hence, $L$ must be a straight line.
• A straight line contains infinitely many elements.

Thus, Option (D) is TRUE.

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Conclusion:

The correct statements are A, C, and D.