A problem regarding complex variable equation $sz+t\bar{z}+r=0$ Let $s,t,r$ be non-zero complex numbers and $L$ be the set of solutions $z=x+iy$ $(x,y \in \mathbb{R}, i^2=-1)$ of the equation 
$$sz+t\bar{z}+r=0$$
where $\bar{z}=x-iy$.Then, which of the following statement(s) is (are) TRUE?
$$\begin{aligned}
\text{(A)}&  \textrm{ If }L \textrm{ has exactly one element, then }|s|\neq |t|\\
\text{(B)}&  \textrm{ If }|s|=|t|,\textrm{ then }L\textrm{ has infinitely many elements}\\
\text{(C)}&  \textrm{ The number of elements in }L \cap \lbrace z:|z-1+i|=5\rbrace\textrm{ is at most } 2\\
\text{(D)}&  \textrm{ If }L\textrm{ has more than one element, then }L\textrm{ has infinitely many elements }\dots
\end{aligned}$$
Can anyone please give me some clue to proceed?
 A: Options (A), (C), (D) are correct:
The relation $$s\bar z+t z+r=0~~~(1) \implies \bar s z + t \bar z +\bar r=0~~~~(2)$$ where $s,t,r$ are complex numbers, being a linear (first degree in $z$) can at best represent a line in Argand plane otherwise a point or nothing (null set).
Eliminating $z$ like in previous Answer, we have
$$(|s|^2-|t|^2)z=\bar r t - r \bar s $$
(A) So when $(|s|^2-|t|^2)\ne 0$, then there is a unique solution: $z=\frac{\bar r t - r \bar s} {(|s|^2-|t|^2)}.$
(B) When $(|s|^2-|t|^2)= 0$, then $\bar r t - r\bar s$ may or may not be zero, so there may be no solution or many solution. In this regard (B) is not essentially true.
(C) The given new locus is a circle and $L$ contains no point, one point or more than one points (line). As a line can cut the circle in at most two points the option (C) is correct.
(D) If $L$ has more than one point it will have infinitely many point as (1) would represent a line 
A: Note
$$sz+t\bar{z}=-r, \>\>\>\>\> \bar s \bar z+\bar t {z}=-\bar r$$
Eliminate $\bar z$ to get
$$(|s|^2 -|t^2|)z = \bar rt-r\bar s$$
Note that, if $|s|^2 \ne |t^2|$, a unique solution $z = \frac{\bar rt-r\bar s}{|s|^2 -|t^2|}$ exists.
On the other hand, if $|s|^2 =|t^2|$, no solutions exist in general. Yet, in the special case where $|s|^2 =|t^2|$ and $\bar rt=r\bar s$, any $z$ satisfies the equation.
Thus, (A) and (D) are true.
