If a complex number has infinitely many solutions why is it a line? Let $s,t,r$ be non-zero complex numbers and $L$ be the set of solutions $z = x+iy \text{ },\:x,y \in \mathbb{R}$ of the equation $sz+t\bar{z}+r=0$ where $\bar{z} = x-iy$.
Prove the number of elements in $L \cap \{z: |z-1+i|=5\}$ is at most $2$.

What they did in the solution is say $z(|s|^2-|t|^2) = \bar{r}t-r\bar{s}$. They did this by taking conjugate of the original equation and eliminating $\bar{z}.$
Now how is $L \cap \{z: |z-1+i|=5\}$ is at most $2$?
If $|s| \neq |t|$, $z = \dfrac{\bar{r}t-r\bar{s}}{(|s|^2-|t|^2)}$ which is one solution and it can intersect the circle once. If $|s| = |t|$ and  $\bar{r}t-r\bar{s}\neq0$ then $L$ is empty. But if $|s| = |t|$ and  $\bar{r}t-r\bar{s}=0$, then $z\cdot0=0 \implies$ $z$ has infinitely many solutions. Why is $L$ a line in this case? Isn't it the entire plane? Why does it intersect the circle at most twice?
Here Z Ahmed told its a line but I can't seem to wrap my head around why this is the case.
 A: Brian seems to have given a complete solution, but to address your specific confusion: You're right that the equation $$z(|s|^2-|t|^2) = \bar{r}t - r\bar{s}$$ will have infinitely many solutions if $|s|^2 - |t|^2 = 0$ and $\bar{r}t - r\bar{s}= 0.$  But, in that case, the elimination method step (where we solve for $z$ by eliminating the conjugate) is producing the trivial equation $0 = 0,$ which loses the original solution set.  The arbitrary $z$ values you choose won't satisfy the original equation - only this secondary equation.
For example, take $s = t = \frac{1}{2}$ and $r = -1$ so that the original equation becomes $$\frac{z}{2} + \frac{\bar{z}}{2} - 1 = 0,$$ which is equivalent to $\Re(z) = 1$.  Clearly that solution set is a line.
If you take the conjugate equation and use it to eliminate $\bar{z}$, then $z$ will be eliminated as well, and you'll get the equation $0z + 0\bar{z} = 0$, which does have infinitely many solutions but isn't equivalent to the original equation.  We've lost the original equation's information.
The overall problem is arising because the conjugate equation is a multiple of the original equation in this degenerate case, so it's not useful for elimination.  For example, if you start with the system of equations \begin{array}{11}x + y = 0\\2x + 2y = 0,\end{array} then there's a line (not plane) of solutions: $\{(x, -x): x \in \mathbb{R}\}$, but the elimination method will kill both variables and lose this information, because this is a dependent system of equations.  You've identified the conditions under which the conjugate trick/elimination method fails.
A: The equation $sz + t\bar{z} + r = 0$ can be rewritten in terms of $x,y$ as
$$s(x+iy) + t(x-iy) + r = 0 \\ (s+t)x + i(s-t)y + r = 0$$

*

*If $s-t \neq 0$, we may solve for $y$ as $$y = \frac{i(s+t)}{s-t}x + \frac{ir}{s-t}$$
which describes either a line, a point, or an empty set.


*If $s-t = 0$, then the equation becomes $$2sx + r = 0 \\ x = -\frac{r}{2s}$$
which describes either a line or an empty set.

Alternatively, if you're having trouble seeing how these are the solution sets, it may be easier to just go back to $$(s+t)x + i(s-t)y + r = 0 \\ \iff \\ \begin{cases}\Re(s+t) x + \Im(t-s)y + \Re(r) = 0 \\ \Im(s+t)x + \Re(s-t)y + \Im(r) = 0\end{cases}$$
so the solution set is the intersection of two real linear equations, each of which is either empty, a line, or the whole plane.  The intersection of two such sets is either empty, a point, a line, or the whole plane, and we may ignore the whole plane as a possibility since, e.g., $r\neq 0$.
A: Your equation is never true on the entire plane (as you've stated the hypothesis that the coefficients $s,t,r$ are nonzero), otherwise, you've defined the function $f(z) = z^{*}$ (* denotes conjugate) as a linear function $z$, which is absurd, not least because $z^{*}$ isn't holomorphic.
You've incorrectly stated -- given a simultaneous system, the equation yielded by subtracting one equation from another provides a complete description of the solution set, but clearly it just gives a new constraint. (well, to be exact not new, just incomplete) Therefore, in the case $|s|=|t|$, you found this new constraint tells you nothing new at all, which just means you have to go back to the original system and solve it some other way.
Perhaps the easiest way to see the solution is a point or a line is to expand the equation into its real and imaginary parts, i.e. you get two linear equations in $x$,$y$. Their solutions are lines. It turns out when $|s|=|t|$, those two lines coincide, and when $|s|\neq |t|$, they do not, therefore those two lines are parallel or intersect at a point. In fact, you've shown they must intersect at a point. To see why $|s|=|t|$ is the right condition for those two lines to coincide, consider a geometric description of a line in the Argand diagram.
