= Solution
The intended open region lies inside the larger circle and outside the smaller one. The two circles are tangent at zero, a boundary point, so $w=1/z$ is holomorphic throughout the region. For a circle $|z-ia|=a$, its equation is $|z|^2=2a\operatorname{Im}z$, which becomes $\operatorname{Im}w=-1/(2a)$. The side inside that circle becomes $\operatorname{Im}w<-1/(2a)$. Hence the <tangent-circle crescent conformal map> begins with
$$
\Omega\longrightarrow\left\{w:-\frac12<\operatorname{Im}w<-\frac14\right\},\qquad w=\frac1z.
$$
The affine coordinate $\zeta=4w+2i$ sends this strip bijectively to $0<\operatorname{Im}\zeta<1$. Exponentiation $\eta=e^{\pi\zeta}$ is then a conformal bijection to the <upper half-plane>: its argument ranges strictly from zero to $\pi$, and the inverse logarithm has that unique argument. Finally $(\eta-i)/(\eta+i)$ is a <Cayley transform between the half-plane and disk> onto the <unit disc>, since for $\operatorname{Im}\eta>0$ its modulus is less than one and the inverse is $\eta=i(1+u)/(1-u)$. As $e^{\pi\zeta}=e^{4\pi/z}$, a complete answer is
$$
\boxed{F(z)=\frac{e^{4\pi/z}-i}{e^{4\pi/z}+i}.}
$$
Every stage has nonzero derivative and is one-to-one onto its indicated range. The denominator cannot vanish in the <upper half-plane>. Unlike a region between disjoint nested circles, this tangent crescent has no enclosed hole, so there is no annulus obstruction to mapping it onto a disc.
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