If I have a complex number z, and I know that 1<|z-1|<2, and Im(z) < 0, I can figure out where the z-1 can lie on the complex plane. But to figure out where z can lie on the complex plane how would I go about doing this?
From 1<|z-1|<2 I can't add one all around cause it would be 2<|z-1|+1<3.
Any help is appreciated.
On
You do add $1$ all around – to each point of the region where $z-1$ lies. This is a translation of the region by the vector (complex number) $1$, and so in this case the centre of the annulus that is the region should lie on $1$.
Adding the condition $\operatorname{Im} z<0$ means clipping to the lower half-plane.
That region (I mean, $\{z\in\mathbb C\mid1<\lvert z-1\rvert<2\}$) is an annulus, that is, the region between two conventric circles (centered at $0$). If you add the condition that $\operatorname{Im}z<0$, what you get is its bottom half.