Let $s,t,r$ be non-zero complex numbers and $L$ be the set of solutions $\{(x,y) \in \mathbb R \}$ and $i=\sqrt{-1}$ of the equation :
$$sz+t\bar z+r=0$$
where $\bar z=x-iy$ prove that it has exactly one element then $|s|\ne |t|$.
I have prove it using $s=a+ib$, $t=c+id$ and $r=e+if$
Using the equation $sz+t\bar z+r=0$, we get two line equation
(a+c)x+(d-b)y+e=0
(b+d)x+(a-c)y+f=0
For unique solution we have $a^2+b^2 \ne c^2+d^2$ viz.$|s|\ne |t|$(neglecting the negative value)
Can I prove it without using the substituion as outlined above
Write the equation as \begin{cases} sz+t\bar{z}=-r \\[4px] \bar{t}z+\bar{s}\bar{z}=-\bar{r} \end{cases} Considering $z$ and $\bar{z}$ as independent variables, we know that the system has a unique solution if and only if $$ \det\begin{bmatrix} s & t \\ \bar{t} & \bar{s} \end{bmatrix}=|s|^2-|t|^2\ne0 $$