1B. Determine the complex numbers $z$ for which

$$
|z|=\frac{1}{|z|}=|z-1|
$$
Solution. It is clear that the equations are defined for $z \neq 0$. From the equation $|z|=\frac{1}{|z|}$, we get $|z|^{2}=1$, hence

$$
|z|=1
$$

From the previous equation and the equation $|z|=|z-1|$, we obtain the equation

$$
|z-1|=1
$$

If we write the complex number $z$ in algebraic form $z=x+i y$, from (1) and (2) we get

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=1 \\
(x-1)^{2}+y^{2}=1
\end{array}\right.
$$

If we subtract the second equation from the first equation, we get the equation $2 x-1=0$. Hence, $x=\frac{1}{2}$. If we substitute this into any of the equations in the system (3), we get the equation $y^{2}=\frac{3}{4}$. Its solutions are $y= \pm \frac{\sqrt{3}}{2}$. Therefore, the set of numbers $\left\{\frac{1}{2}+i \frac{\sqrt{3}}{2}, \frac{1}{2}-i \frac{\sqrt{3}}{2}\right\}$ is the solution to the equations.