How to graph $|z-1| <2$
By John Peck •
Am I correct to rearrange this to $(z-1)^2 < 4$, and hence just graph as a circle or am I completely off?
$\endgroup$ 11 Answer
$\begingroup$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 $2$.
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