Question

`|z-1-i| < 1`?

Answer 1

This is the interior of a circular disc of radius `1` centred on `1+i`

Explanation:

Note that `abs(z)` is the size of `z`, regardless of its direction.

So `abs(z-a) = r` is the set of points at distance `r` from `a`.

So `abs(z-1-i) = 1` is a circle of radius `1` with centre `1+i`.

So the inequality `abs(z-1-i) < 1` is the set of points at distance less than `1` from `1+i`. In other words, it is the interior of the circle of radius `1` with centre `1+i`

graph{(x-1)^2+(y-1)^2 < 1 [-1.843, 3.157, -0.24, 2.26]}