I need to sketch this region $\left \{ z\in\mathbb{C}| |z-i|\leq |z-1| \right \}$. I'd like some assistance with solving this inequality because I think that's where I'm going wrong.
To solve the inequality I'm squaring both sides and trying to solve for that. Similar to this post. $$(z-i)^2 \leq (z-1)^2$$ $$0\leq (z-1)^2 - (z-i)^2$$ $$0\leq ((z-1) - (z-i)) ((z-1) + (z-i))$$ $$0\leq (-1+i)(2z-1-i)$$ $$0\leq -2z+2zi+-i^2+1$$
Here is the point where I get stuck. I'm not quite sure how to progress from here.
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$\begingroup$note that : $z=x+iy \rightarrow \space |z|=\sqrt{x^2+y^2}$ $$|z-i|<|z-1|\\$$put $z=x+iy$ $$ |x+iy-i|<|x+iy-1|\\|x+i(y-1)|<|(x-1)+iy|\\\sqrt{x^2+(y-1)^2}<\sqrt{(x-1)^2+y^2}$$ now go on
when you simplify
$$x^2+(y-1)^2 <(x-1)^2+y^2\\-2y<-2x\\2x<2y\\x<y$$
The geometric interpretation of your inequality is:
The distance from $z$ to $i$ is less than or equal to the distance from $z$ to $1$.
Draw the perpendicular bisector to the segment from $1$ to $i$ and think about what this means.
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