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!

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.)
