Algebraic trick to map $|z|<2$ Suppose that we want to find the image of the region $|z|<1$ under the mapping $w=\frac z{z+1}$. Since $z=\frac{-w}{w-1}$ (and assuming $w= u+iv$) we should have $\left|\frac w{w-1}\right|=\left|\frac{u(u-1)+v^2-iv}{(u-1)^2+v^2}\right|<1$ or equivalently
$$u^2(u-1)^2+v^4+2v^2u(u-1)+v^2<(u-1)^4+v^4+2v^2(u-1)^2.$$
Now we can use the following algebraic trick to write
$$(u-1+1)^2(u-1)^2+2v^2(u-1+1)(u-1)+v^2<(u-1)^4+2v^2(u-1)^2$$
which implies that
$$(u-1)^2+2(u-1)(u-1)^2+2v^2(u-1)+v^2<0$$
or
$$((u-1)^2+v^2))(2u-1)<0.$$
Thus the image would be $2u-1<0$.
Now I wonder if there is some similar algebraic trick for $|z|<2$. In this case we have
$$u^2(u-1)^2+v^4+2v^2u(u-1)+v^2<4(u-1)^4+4v^4+8v^2(u-1)^2$$
I've tried similar method but couldn't arrive to something useful. Could anyone help me in this case? Thanks!
 A: The function $f(z)={z\over1+z}$ is a Möbius transformation, so it takes lines and circles to lines and circles.  Clearly, it takes the real axis to the real axis.  Since $f(-1)=\infty,$ f must take the circle $|z|=2$ to a circle.  Now, Möbius transformations are conformal, so as the real axis and the circle $|z|=2$ are orthogonal, their images must be orthgonal, which is to say that the center of $f(|z|=2)$ must lie on the real axis.  As $f(2)=\frac23, f(-2)=2$ we see that the image of $|z|=2$ is $|z-4/3|=2/3.$  
We now know that $f$ maps $|z|<2$ either to the interior or the exterior of $|z-4/3|=2/3.$  Since $f(0) = 0,$ it must be the exterior.  Thus the image is the region $|z-4/3|>2/3.$     
A: The same story with different words. Write the function this way:
$$ f(z) = {z+1-1\over z+1} = 1- {1\over z+1}$$ 
Now this function is composition of following functions $$z\mapsto z+1,$$ $$z\mapsto {1\over z},$$ $$z\mapsto -z$$ and finally $$z\mapsto 1+z$$ 
First one moves region $R_1:\; |z|<1$ for one to the left so we get $R_2:\; |z+1|<2$. Second one inverts $R_2$ accros $0$ to $R_3:\; |z+{1\over 3}|>{2\over 3}$. Third transformation just relect this region into/onto it self, and finaly last transformation just move $R_3$ to the right for 1 into $|z-{2\over 3}|>{2\over 3}$  
A: For a direct algebraic proof, start the same way as in the first case with $\,z = \dfrac{w}{1-w}\,$, then:
$$
\begin{align}
|z| \lt 2 \quad&\iff\quad |w|^2 \lt 4 \cdot|1-w|^2 \\
 &\iff\quad w \bar w \lt 4 \cdot(1-w)(1-\bar w) \\
 &\iff\quad 3 w \bar w - 4 w - 4 \bar w \color{red}{+\frac{16}{3}-\frac{16}{3}} + 4 \gt 0 \\
 &\iff\quad 3\left(w-\frac{4}{3}\right)\left(\bar w-\frac{4}{3}\right) \gt \frac{4}{3} \\
 &\iff\quad \left|w-\frac{4}{3}\right|^2 \gt \frac{4}{9} \\[5px]
 &\iff\quad \left|w-\frac{4}{3}\right| \gt \frac{2}{3}
\end{align}
$$
Therefore the image is the exterior of the circle of radius $\,2/3\,$ centered at $\,4/3\,$.
(Geometrically, the locus of points in the plane with constant ratio $\,\left|\frac{w}{w-1}\right| = 2\,$ between the distances to the two fixed points $\,0, 1\,$ is a circle of Apollonius having as diameter the points that divide the segment between $\,0\,$ and $\,1\,$ in ratio $\,2:1\,$ internally and respectively externally. In this case the endpoints of that diameter are $\,2/3$ and $\,2\,$, so the circle is centered at $\,(2+2/3)/2=4/3\,$ and has a radius $\,(2-2/3)/2=2/3\,$, which is of course the same as found above.)
