How to graph $|z-1| <2$

Question

Am I correct to rearrange this to $(z-1)^2 < 4$, and hence just graph as a circle or am I completely off?

Answer 1

Assume $z=x+iy$, where $x$ and $y$ are real numbers. Notice that the inequality $|z-1|<2$ can be written as $$\sqrt{(x-1)^2+y^2}<2$$ Squaring both sides yields $$(x-1)^2+y^2<4$$

which is the set of points inside (and not on) the circle centered at $(1,0)$ with radius $2$.