How to plot $\{z \in \mathbb{C} : |z-i|>|z+i|\}$ 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 $\{z \in \mathbb{C} : |z-i|>|z+i|\}$


 A: Write out a complex number $z$ with real and imaginary components, i.e. as $z=a+bi$. Then
$$|z-i|=|a+(b-1)i|=\sqrt{a^2+(b-1)^2}$$
$$|z+i|=|a+(b+1)i|=\sqrt{a^2+(b+1)^2}$$
so
$$\begin{align*}|z-i|>|z+i|&\iff\sqrt{a^2+(b-1)^2}>\sqrt{a^2+(b+1)^2}\\&\iff a^2+(b-1)^2>a^2+(b+1)^2\\ &\iff (b-1)^2>(b+1)^2\\ &\iff -2b>2b\\ &\iff b<0\end{align*}$$
Thus, the complex number $z=a+bi$ satisfies $|z-i|>|z+i|$ if and only if $b<0$, i.e. if and only if it lies below the real axis in the complex plane. Thinking geometrically (i.e. with complex numbers as points in the plane), it might also help to note that
$$|z-i|=\sqrt{a^2+(b-1)^2}=\text{distance from }(a,b)\text{ to }(0,1)$$
$$|z+i|=\sqrt{a^2+(b+1)^2}=\text{distance from }(a,b)\text{ to }(0,-1)$$
A: Try writing $z=x+iy$ with $x,y\in\mathbb{R}$.
Then for the first inequality you get (just try it for yourself):
$$|z-i|>|z+i| \Leftrightarrow \mbox{Im}(z) < 0$$
so the solutions is the whole lower half-plane (without the real axis).
For the second one, you get $\mathbb{C}\setminus\mathbb{R}$, because of a similar condition for the imaginary part ($\mbox{Im}(z)\not=0$).
Also, your last line $i(z-\bar{z})>i(\bar{z}-z)$ translates for $z=x+iy$ into $0>y$, which gives the same result.
A: One could simply look at the absolute value of a complex number as it's distance from origin, or in other words, $|a-b|$ is the distance of $a$ from $b$. 
Now, in your problem you want find all the points such that their distance from $i$ is more than their distance from $-i$. Well, find all the points who have the same distance from $i$ and $-i$. That's a line (namely the line that is perpendicular to the line segment $\overline{i,-i}$). 
So far you've got the solution to the second part of the problem. Now, choose the half-plane which includes $-i$. That is that answer for the first part.
