 Solution

The two bounding circles meet at zero and one, which the Möbius transformation sends to zero and infinity. They therefore become straight lines through zero. More explicitly write $w=f(z)$, so $z=w/(1+w)$. The two disk inequalities becomeSince $\zeta =1/2+i\sqrt{3}/2$, these say $\mathrm{Re}w>\sqrt{3}|\mathrm{Im}w|$. ThusThe positive real axis belongs to the image because $f(1/2)=1$, fixing which sector is intended. Cubing maps this sector bijectively onto the right half-plane: its argument interval of width $\pi /3$ becomes $(-\pi /2,\pi /2)$. The Cayley transform between the half-plane and disk $v\mapsto (v-1)/(v+1)$ then maps that half-plane bijectively onto the unit disk. Hence a holomorphic bijection isEvery denominator is nonzero on the corresponding domain, and the sector width ensures the power map is injective. This is the conformal map of a symmetric circular lens.