Let $s,t,r$ be non zero complex numbers and $L$ is is set of solutions of $z=x+iy$ of the equation $sz + t\bar z+r=0$ 
Prove that $L$ is a singleton set if $|s|\not =|t|$


And, prove that $z$ is a straight line if $L$ is not singleton

Solving the equation, I got $z=\frac{\bar s r -\bar r t}{|t|^2-|s|^2}$
I personally cannot see any reason why $z$ will have a unique solution if $|s|\not = |t|$, because then $z=k(\bar s r -\bar r t)$
I have no idea whether this represents a line or a point because I have serious conceptual problems with complex numbers, which I hope to clear. I know a question similar to this exists on MSE, but none of the answers justify their claims.
 A: You proved that $z=\frac{\overline s r - \overline r t}{|t|^2-|s|^2}$. Then you claim that

$|s|\not = |t|$, because then $z=k(\bar s r -\bar r t)$

What do you mean by this?
In fact, your equality $z=\frac{\overline s r - \overline r t}{|t|^2-|s|^2}$ already proves that if $|t|\neq |s|$, then $z$ can only have one single value, i.e. the value $\frac{\overline s r - \overline r t}{|t|^2-|s|^2}$.
A: The equation $sz+t\bar{z}=-r$, when written in real coordinates with $s=s_1+is_2$, $t=t_1+it_2$, $r=r_1+ir_2$, and $z=x+iy$, becomes two real equations $$(s_1+t_1) x - (s_2- t_2) y= -r_1,\qquad (s_2+ t_2) x + (s_1- t_1) y = -r_2$$ In matrix form, $$\begin{pmatrix}s_1+t_2&t_2-s_2\\s_2+t_2&s_1-t_1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=-\begin{pmatrix}r_1\\r_2\end{pmatrix}$$
Normally this would give one solution for $(x,y)$ and hence for $z$ as in the question. But if $|s|=|t|$ then $s_1^2+s_2^2=t_1^2+t_2^2$, so the above two equations reduce to just one (their determinant is zero). The second equation becomes a multiple of the first. One equation in $x,y$ represents a straight line in the complex plane.
However it is important to realise that the RHS of the second equation must be compatible ($r_1(s_1-t_1)=r_2(t_2-s_2)$), otherwise there are no solutions.
