Show that $|\frac{z}{1-z}|<1 \Leftrightarrow \operatorname{Re}z < \frac{1}{2}$ Let $z \neq 1$ be a complex number. Show that $$\left|\frac{z}{1-z}\right|<1 \Leftrightarrow \operatorname{Re}z < \frac{1}{2}$$
I have rewritten it as $\frac{\sqrt{a^2+b^2}}{\sqrt{(1-a)^2+b^2}}$, but not sure where to go from here.

You're on the right track, though it's better to work with the absolute value squared. If $z=a+bi$ and $|z|^2<|z-1|^2$ then 
$$ a^2+b^2<(a-1)^2+b^2$$
which implies that
$$ 2a-1<0$$
or $a<\frac{1}{2}$. By reversing the argument above, if $a<\frac{1}{2}$ then $|z|^2<|z-1|^2$.
