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.

  • $\begingroup$ There's one variable $z=x+iy$ with $2$ real unknowns; and there is one linear complex equation, which is $2$ real equations. Doesn't that indicate one solution at most (generically)? $\endgroup$ – Chrystomath Sep 16 at 6:51
  • $\begingroup$ @Chrystomath $2$ real unknowns and $2$ equations can have an infinitely many solutions. See the equations $0=0$ and $0=0$. Or, less trivially, the equations $x+y=1$ and $2x+2y=2$. $\endgroup$ – 5xum Sep 16 at 6:52
  • $\begingroup$ @5xum Yes of course, as in all linear equations. That case is covered by $|t|=|s|$. $\endgroup$ – Chrystomath Sep 16 at 6:53
  • $\begingroup$ @Chrystomath So doesn't that answer your original question? $\endgroup$ – 5xum Sep 16 at 6:56
  • $\begingroup$ @5xum I'm addressing the asker. It's their question: "I personally don't see why $z$ will have a unique solution". $\endgroup$ – Chrystomath Sep 16 at 7:01

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 $\frac{\overline s r - \overline r t}{|t|^2-|s|^2}$.

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  • $\begingroup$ The asker is not querying whether $z=\cdots$ represents one solution. They're asking whether the solution should be a line rather than a point (a priori), presumably because the equation looks linear of the sort $y=mx+c$. $\endgroup$ – Chrystomath Sep 16 at 7:04
  • $\begingroup$ @Chrystomath Maybe. Maybe not. I'd rather the OP explain what he means himself. $\endgroup$ – 5xum Sep 16 at 7:05
  • $\begingroup$ Yes, what Chrystomath said was right. I don’t know why it represents a point, since the very next question asks to prove it as a line $\endgroup$ – Aditya Sep 16 at 7:08
  • $\begingroup$ @Aditya The equation represents a point if $|t|\neq |s|$. You have shown this by explicitly finding the point represented by the equation. The equation represents a line if $|t|=|s|$. This is not contradictory, as you can see the equation as a set of two real equations for two real variables. $\endgroup$ – 5xum Sep 16 at 7:09
  • $\begingroup$ If $|t|=|s|$, then isn’t $z=\infty$? How is that a line? $\endgroup$ – Aditya Sep 16 at 7:13

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.

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  • $\begingroup$ What about going by my method? $\endgroup$ – Aditya Sep 16 at 8:49

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