A problem regarding complex variable equation $sz+t\bar{z}+r=0$

Question

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?

Answer 1

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.

Answer 2

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