## Task A-2.1.

Determine all real numbers $a$ such that there exists a complex number $z$ for which

$$
|z|=1 \quad \text { and } \quad|a z-1|=a|z+1|
$$
## First solution.

If $a > 0$, then from the second equality we have

$$
|a z - 1| = |a z + a|
$$

Since $|w|^2 = w \cdot \bar{w}$ for $w \in \mathbb{C}$, the equality $(*)$ is equivalent to

$$
\begin{gathered}
(a z - 1)(a \bar{z} - 1) = (a z + a)(a \bar{z} + a) \\
a^2 z \bar{z} - a z - a \bar{z} + 1 = a^2 z \bar{z} + a^2 z + a^2 \bar{z} + a^2 \\
(a^2 + a)(z + \bar{z}) = 1 - a^2
\end{gathered}
$$

From this, we get $z + \bar{z} = \frac{1 - a^2}{a(a + 1)}$, since $a \neq 0, a \neq -1$.

1 point

Since $z + \bar{z} = 2 \operatorname{Re} z$, we conclude that $\operatorname{Re} z = \frac{1 - a^2}{2 a(a + 1)} = \frac{1 - a}{2 a}$.

For such a complex number $z$ to exist, for which $|z| = 1$ also holds, it is necessary and sufficient that $-1 \leq \operatorname{Re} z \leq 1$, i.e.,

$$
-1 \leq \frac{1 - a}{2 a} \leq 1
$$

From this, due to $a > 0$, we get $a \geq \frac{1}{3}$.

Finally, the desired numbers are all $a \in \{-1\} \cup \left[\frac{1}{3}, \infty\right)$.