Region of $|z^2-2z+2|<2$

Question

I am getting trouble with finding the region of $$|z^2-2z+2|<2$$

I have tried the way $z=x+iy$, but I got lost with alot of terms, so I really would appreciate any help with that

Answer 1

Let $z=x+iy$ to rewrite $|z^2-2z+2|^2<4$ as

$$[(x+i y)^2-2(x+i y)+2][(x-i y)^2-2(x-i y)+2]-4<0$$

Simplify to get

$$(y^2+x^2-2x)^2+4(x^2-2x)<0$$

which represents the shaded region in the graph.

Answer 2

Also

$$2>|z^2-2z+2|=|(z-1)^2+1||> |(z-1)^2|-|1|\implies |z-1|^2<3 $$ $$\implies 1-\sqrt{3}<|z-1| <1+\sqrt{3} \implies |z-1| < 1+\sqrt{3}.$$ So the required region is bounded and it ilies in the circle of radius $1+\sqrt{3}$ centered at $(1,0)$