 Solution

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 $w\ne 0$ the first inequality becomesThe second becomesThus the image is contained in the vertical strip $1/4<\mathrm{Re}w<1/2$. Conversely every point in that strip is nonzero; its inverse satisfies both original inequalities, so no points are missing. ThereforeThe 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.