How would I draw the set $\{z \in \mathbb{C} : |z-i|>|z+i|\}$ and $\{z \in \mathbb{C} : |z-i|\not=|z+i|\}$?
Im not sure how to solve the second one, and for the first one, I tried squaring both sides and trying to work something out, but I got no where.
$|z-i|^2>|z+i|^2\\\\(z-i)(\bar z+i)>(z+i)(\bar z-i)\\ z\bar z+1+i(z -\bar z)>z \bar z+1 +i(\bar z -z)\\i(z-\bar z)>i(\bar z-z)$
What would the 'general' method/approach be for drawing the sets?
Edit: How would I draw $\{z \in \mathbb{C} : |z-i|\not=|z+i|\}$?
After a similar calculation using Zev Chonoles' post, I got that $-b\not=b$, hence $z=a+ib$ satisfies $|z-i|\not=|z+i|$ if and only if $-b\not=b$.
For $\{z \in \mathbb{C} : |z-i|>|z+i|\}$