1. Find all complex numbers $z$ for which

$$
\left|\frac{1}{z-i}+1\right|=1 \quad \text { and } \quad\left|\frac{1}{z-i}-i\right|=1
$$
1. Solution. From the given condition, it is clear that

$$
\frac{1}{z-i}=0 \quad \text { or } \quad \frac{1}{z-i}=-1+i
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_db59002be042233d39ccg-05.jpg?height=295&width=399&top_left_y=570&top_left_x=908)

The first case is obviously not possible.

5 points

In the second case,

$$
\begin{aligned}
\frac{1}{z-i} & =-1+i \\
z-i & =\frac{1}{-1+i} \cdot \frac{-1-i}{-1-i}=-\frac{1}{2}-\frac{i}{2} \\
z & =-\frac{1}{2}+\frac{i}{2}
\end{aligned}
$$

15 points

Second solution. Substituting \( z=x+i y \), we get

$$
\begin{aligned}
& \left|\frac{1}{x+i y-i}+1\right|=1 \\
& \left|\frac{1}{x+i y-i}-i\right|=1, \quad 5 \text { points } \\
& \left|\frac{x+1+i(y-1)}{x+i(y-1)}\right|=1 \\
& \left|\frac{y-i x}{x+i(y-1)}\right|=1 \\
& |x+1+i(y-1)|=|x+i(y-1)| \\
& |y-i x|=|x+i(y-1)| \\
& (x+1)^{2}+(y-1)^{2}=x^{2}+(y-1)^{2} \\
& y^{2}+x^{2}=x^{2}+(y-1)^{2} \\
& x=-\frac{1}{2} \\
& y=\frac{1}{2} \\
& 15 \text { points }
\end{aligned}
$$