Determine the set of all complex numbers $\displaystyle z \in \mathbb {C}$ for which $\displaystyle |z+1| < 2|z-i|$ holds, and draw a figure of that set.

So I write up the inequality using $\displaystyle z=a+bi$ which gives

$\displaystyle \sqrt{a^2 +2a + 1 + b^2} < 2 \sqrt {a^2 + b^2 -2b +1 }$ but don't really know how to proceed from there. I seem to find it confusing that I end up with an inequality in both a's and b's.