1. ## Complex Analysis

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

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

$\sqrt{a^2 +2a + 1 + b^2} < 2 \sqrt {a^2 + b^2 -2b +1 }$ 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.

2. Originally Posted by nmatthies1
Determine the set of all complex numbers $z \in \mathbb {C}$ for which $|z+1| < 2|z-i|$ holds, and draw a figure of that set.

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

$sqrt{a^2 +2a + 1 + b^2} < 2 sqrt {a^2 + b^2 -2b +1 }$ 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.
Determine the locus defined by |z + 1| = 2 |z - i| (it's a Circle of Apollonius). Now test the interior and exterior of this circle - which region satisfies the given inequality.

3. Originally Posted by nmatthies1
Determine the set of all complex numbers $z \in \mathbb {C}$ for which $|z+1| < 2|z-i|$ holds, and draw a figure of that set.

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

$\sqrt{a^2 +2a + 1 + b^2} < 2 \sqrt {a^2 + b^2 -2b +1 }$ 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.
Square both sides, gather together simmilar terms on the same side (I'd choose on the right side), complete squares and check this is the exterior of a

circle of radius $\frac{2\sqrt{2}}{3}$

Tonio