4. Determine the set of points $z$ in the complex plane for which there exists a real number $c$ such that $z=\frac{c-i}{2 c-i}$.
Solution. Let $z=x+i y$. We have:

$$
c=\frac{z i-i}{2 z-1}=\frac{-y+(x-1) i}{2 x-1+2 y i} \cdot \frac{2 x-1-2 y i}{2 x-1-2 y i}=\frac{(1-2 x) y+(x-1) 2 y+\left[(x-1)(2 x-1)+2 y^{2}\right] i}{(2 x-1)^{2}+4 y^{2}}
$$

The resulting set is

$$
\begin{aligned}
S & =\left\{z=x+i y \mid(x-1)(2 x-1)+2 y^{2}=0,(x, y) \neq\left(\frac{1}{2}, 0\right)\right\} \\
& =\left\{z=x+i y \left\lvert\,\left(x-\frac{3}{4}\right)^{2}+y^{2}=\frac{1}{16}\right.,(x, y) \neq\left(\frac{1}{2}, 0\right)\right\}
\end{aligned}
$$

## Junior Olympiad