Let `s ,\ t ,\ r` be non-zero complex numbers and `L` be the set of solutions `z=x+i y\ \ (x ,\ y in RR,\ \ i=sqrt(-1))` of the equation `s z+t z +r=0` , where ` z =x-i y` . Then, which of the following statement(s) is (are) TRUE? If `L` has exactly one element, then `|s|!=|t|` (b) If `|s|=|t|` , then `L` has infinitely many elements (c) The number of elements in `Lnn{z :|z-1+i|=5}` is at most 2 (d) If `L` has more than one element, then `L` has infinitely many elements

Options

A. If L has exactly one element, then `|s| ne |t|`
B. If `|s| = |t|` then L has infinitely many elements
C. The number of elements in `L nn {z :|z-1+i|=5}` is at most 2
D. If L has most than one elements, then L has infinitely many elements.

Explanation

To solve the problem, we need to analyze the given equation and the conditions for the set of solutions \( L \). The equation given is:

\[ s z + t z + r = 0 \]

where \( z = x + i y \) and \( x, y \in \mathbb{R} \).

### Step 1: Rewrite the equation
First, we can rewrite the equation as:

\[ (s + t) z + r = 0 \]

This implies:

\[ (s + t) z = -r \]

### Step 2: Solve for \( z \)
Now, we can express \( z \):

\[ z = \frac{-r}{s + t} \]

### Step 3: Analyze the solutions
1. If \( s + t \neq 0 \), then there is exactly one solution for \( z \) given by \( z = \frac{-r}{s + t} \).
2. If \( s + t = 0 \), then we have \( s = -t \). In this case, the equation becomes:

\[ r = 0 \]

Since \( r \) is a non-zero complex number, this case is not possible.

### Step 4: Conditions for the number of solutions
We can analyze the conditions given in the options:

- **(a)** If \( L \) has exactly one element, then \( |s| \neq |t| \).
- This statement is **TRUE** because if \( |s| = |t| \), then \( s + t \) could potentially be zero, leading to no unique solution.

- **(b)** If \( |s| = |t| \), then \( L \) has infinitely many elements.
- This statement is **FALSE** because \( |s| = |t| \) does not guarantee that \( r = 0 \). Thus, it can lead to no solutions.

- **(c)** The number of elements in \( L \) such that \( |z - (1 + i)| = 5 \) is at most 2.
- This statement is **TRUE** because a line can intersect a circle at most at two points.

- **(d)** If \( L \) has more than one element, then \( L \) has infinitely many elements.
- This statement is **FALSE** because having more than one solution does not imply infinitely many solutions; it could just be two solutions.

### Conclusion
The true statements are:
- (a) and (c).