# 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.

• 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)? – Chrystomath Sep 16 at 6:51
• @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 $2x+2y=2$. – 5xum Sep 16 at 6:52
• @5xum Yes of course, as in all linear equations. That case is covered by $|t|=|s|$. – Chrystomath Sep 16 at 6:53
• @Chrystomath So doesn't that answer your original question? – 5xum Sep 16 at 6:56
• @5xum I'm addressing the asker. It's their question: "I personally don't see why $z$ will have a unique solution". – 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 $$\frac{\overline s r - \overline r t}{|t|^2-|s|^2}$$.

• 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 $y=mx+c$. – Chrystomath Sep 16 at 7:04
• @Chrystomath Maybe. Maybe not. I'd rather the OP explain what he means himself. – 5xum Sep 16 at 7:05
• Yes, what Chrystomath said was right. I donâ€™t know why it represents a point, since the very next question asks to prove it as a line – Aditya Sep 16 at 7:08
• @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. – 5xum Sep 16 at 7:09
• If $|t|=|s|$, then isnâ€™t $z=\infty$? How is that a line? – 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.

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