I ask for help in the decision. Tried to put z = x + yi , but in the end I do not get the intersection, that is:

im (x + yi + 2i) & lt; 0.
Im (x + (y + 2) i) & lt; 0.
Y + 2 & lt; 0.
Y & LT; -2.

| z - i | & lt; 1
| X + Yi - I | & lt; 1
| X + (Y - 1) I | & lt; 1
SQRT (X ^ 2 + (Y - 1) ^ 2) & lt; 1
x ^ 2 + (y - 1) ^ 2 & lt; 1

As a result, we obtain a circle with a radius 1 (that is, the length of the complex number should be less than 1, enter this area), and from the previous equation it is clear that Y & LT; -2

Now it is not clear whether the solution is true? If so, why, if not, tell me, please, where it was wrong.

Answer 1, Authority 100%

Solutions for this system of inequalities there is no, which shows your schedule
those. On the complex, nothing to portray and do not need an empty sheet 🙂