TITLE: Showing that $\{z\in\mathbb{C}:|z-1|<|z+i|\}$ is an open set
QUESTION [5 upvotes]: Got stuck on some homework (from H. A. Priestley, Complex Analysis). My topology ain't quite up to speed yet.
So, I want to show that $S=\{z\in\mathbb{C}:|z-1|<|z+i|\}$ is open. Geometrically it's the points above the line through $1-i$ and the origin. (Eh?)
So, essentially what I want to do is to prove that for every $z\in S$, there's a $r>0$ such that $D(z;r)\subset S$. In more concrete words, I want to figure out an $r$ such that the implication $|w-z|<r \Rightarrow |w-1|<|w+i|$ holds.
I started to fiddle around with the triangle inequality
$$|w-1|=|w-z+z-1|\leq |w-z|+|z-1|$$
and then figured I could set $0<r<|z+i|-|z-1|$, which then would yield
$$|w-1|\leq |z+i|$$
but that isn't quite it. I tried to use more terms in the "triangle inequality trick" but I can't seem to get it the way I want it. Help me oh math.stackexchange! (If you want, of course.)

REPLY [3 votes]: Choose $r = \frac{|z+i|-|z-1|}2$. Then, if $|w-z|<r$,
$$
|w-1|<|z-1|+r= |z+i|-r<|w+i|.
$$