= Solution
{c}
Let $w=1/(z+1)$ and invert to $z=1/w-1$. The <pole> $z=-1$ lies on the excluded boundary, so it is not a point of the original region. For $w\neq0$ the first inequality becomes
$$
|z|>1
\iff|1-w|>|w|
\iff1-2\operatorname{Re}w>0.
$$
The second becomes
$$
|z-1|<2
\iff|1-2w|<2|w|
\iff1-4\operatorname{Re}w<0.
$$
Thus the image is contained in the vertical strip $1/4<\operatorname{Re}w<1/2$. Conversely every point in that strip is nonzero; its inverse satisfies both original inequalities, so no points are missing. Therefore
$$
\boxed{w(\mathcal R_1)=\{w:\tfrac14<\operatorname{Re}w<\tfrac12\}.}
$$
The two circles pass through the <pole> of the <Möbius transformation>, which is why their images are straight lines. Direct inequality transformation fixes which side of each line is required.
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