= Solution
{c}
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 become
$$
\operatorname{Re}(\overline\zeta w)>0,\qquad\operatorname{Re}(\zeta w)>0.
$$
Since $\zeta=1/2+i\sqrt3/2$, these say $\operatorname{Re}w>\sqrt3|\operatorname{Im}w|$. Thus
$$
\boxed{f(R)=\{w\ne0:-\pi/6<\arg w<\pi/6\}.}
$$
The 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 is
$$
\boxed{h(z)=\frac{\left(\frac z{1-z}\right)^3-1}{\left(\frac z{1-z}\right)^3+1}
=\frac{z^3-(1-z)^3}{z^3+(1-z)^3}.}
$$
Every 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>.
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