Let $s,t,r$ be non-zero complex numbers and $L$ be the set of solutions $z=x+iy \, (x,y \in \mathbb{R}$ of the equation $sz+t\bar{z}+r=0$. Then prove that if $L$ can not have exactly 10 elements.
I calculated $z=\dfrac{t\bar{r}-r\bar{s}}{\mid{s}{\mid}^2-\mid{t}{\mid}^2}$. But how to proceed further?
On
From the given, we have
$$sz+t\bar{z}=-r, \>\>\> \bar s \bar z+\bar t {z}=-\bar r\implies (|s|^2 -|t^2|)z = \bar rt-r\bar s$$
Two cases to consider:
1) If $|s|^2 \ne |t^2|$, there is a unique solution $z = \frac{\bar rt-r\bar s}{|s|^2 -|t^2|}$.
2) If $|s|^2 =|t^2|$, no solutions exist for $\bar rt\ne r\bar s$; or, infinite number of solutions for $\bar rt=r\bar s$.
Thus, the set can not have exactly 10 elements.
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If you write out your equation in terms of the real and imaginary parts of all the complex numbers involved ($s,t,r,z$) and then use the fact that a complex number (like $sz+t\bar z+r$) is $0$ if and only if both its real part and its imaginary part are $0$, you'll get two linear equations for the two unknowns $x$ and $y$. So you can apply the theorem from linear algebra that a system of linear equations (over the real numbers) has either no solutions or a single solution, or infinitely many solutions. In particular, it can't have exactly $10$ solutions.
There is no need to compute anything.
If the equation has more than one solutions, say $z_1$ and $z_2$, then any linear combination of $z_1$ and $z_2$ of the form $\lambda z_1 + (1-\lambda)z_2$ with $\lambda \in \mathbb{R}$ is also a solution.
This means your equation has either $0, 1$ or infinitely many solutions. It is impossible for it to have exactly $10$ solutions.