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.
 A: 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 $|z|^2<|z-1|^2$.
A: I would write (in anticipation of the conclusion) $z=1/2-w$
(so that on one side we have Re$\,w>1/2$). On the other side we have
$$\left|\frac{1/2-w}{1/2+w}\right|<1.$$
Geometrically that says $w$ is nearer to $1/2$ than to $-1/2$.
A: Well, when $\text{z}\in\mathbb{C}$:
$$\left|\frac{\text{z}}{1-\text{z}}\right|=\frac{\left|\text{z}\right|}{\left|1-\text{z}\right|}=\frac{\sqrt{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)}}{\sqrt{\left(1-\Re\left(\text{z}\right)\right)^2+\Im^2\left(\text{z}\right)}}=\sqrt{\frac{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)}{\left(1-\Re\left(\text{z}\right)\right)^2+\Im^2\left(\text{z}\right)}}\tag1$$
So, when:
$$\left|\frac{\text{z}}{1-\text{z}}\right|<1\space\Longleftrightarrow\space\frac{\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)}{\left(1-\Re\left(\text{z}\right)\right)^2+\Im^2\left(\text{z}\right)}<1\space\Longleftrightarrow$$
$$\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)<\left(1-\Re\left(\text{z}\right)\right)^2+\Im^2\left(\text{z}\right)\space\Longleftrightarrow\space\Re^2\left(\text{z}\right)<\left(1-\Re\left(\text{z}\right)\right)^2\space\Longleftrightarrow$$
$$\Re\left(\text{z}\right)<1-\Re\left(\text{z}\right)\space\Longleftrightarrow\space2\cdot\Re\left(\text{z}\right)<1\space\Longleftrightarrow\space\Re\left(\text{z}\right)<\frac{1}{2}\tag2$$
A: if
$$
f(z)=\frac{z}{1-z}
$$
then 
$$
|f(\frac12+yi) = \bigg|\frac{\frac12+iy}{\frac12-yi}\bigg|=1
$$
so the fractional linear transformation $f$ maps the line $x+\frac12$ to the unit circle. The interior of the unit disc is the image of one of the two open half-planes separated by the line. as the origin is a fixed point for $f$, it is the left-hand region $\mathfrak{Re}(z) \lt \frac12$ which maps to the interior.  
A: The inverse of $w=\frac{z}{1-z}$ is $z=\frac{w}{1+w}$. Since $\frac{w}{1+w}$ is a linear fractional transformation, it sends circles and lines to circles and lines. It sends the points $\{-i,1,i\}$ to $\left\{\frac{1-i}2,\frac12,\frac{1+i}2\right\}$. Thus, it sends the unit circle to the line $\operatorname{Re}(z)=\frac12$. Since it sends $0$ to $0$, we get the proper mapping.
A: Hint: $\require{cancel}\;|z|^2 \lt |1-z|^2 \iff \cancel{z \bar z} \lt (1-z)(1-\bar z) = 1 - z - \bar z + \cancel{z \bar z} \iff z + \bar z \lt 1$
