If you have a circle centre $z_0$ radius $R$ (i.e. the points $|z-z_0|=R$), the ''inversion'' in that circle is defined as the map which takes a point $z\ne z_0$ to a point $w$ on the line segment through $z$ and starting at $z_0$ such that the ratio of $R$ to $|z-z_0|$ is the same as the ratio of $|w-z_0|$ to $R$. In other words, $|w-z_0|=R^2/|z-z_0|$. This is called ''inversion in the circle $|z-z_0|=R$'' because the inside of the circle gets mapped to the outside, the outside gets mapped to the inside, and the boundary is mapped to itself. It inverts ''inside'' and ''outside''. If $z_0=0$ and $R=1$, we get $\left|w\right|=1/\left|z\right|$. So we can show that $w=z/|z{|}^2$. If $z=r{e}^{i\theta }$, then $1/z=(1/r)e^{-i\theta }$. Inverting $z$ in the circle $\left|z\right|=1$ gives us $w=(1/r)e^{i\theta }$, reflecting that in the $x$ axis changes the sign of the angle, so we get $(1/r)e^{-i\theta }$, which is just $1/z$.