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# portray on the complex plane set of points satisfying the condition

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Category 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.

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 🙂

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