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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 $|z{|}^2=2a\mathrm{Im}z$, which becomes $\mathrm{Im}w=-1/(2a)$. The side inside that circle becomes $\mathrm{Im}w<-1/(2a)$. Hence the tangent-circle crescent conformal map begins withThe affine coordinate $\zeta =4w+2i$ sends this strip bijectively to $0<\mathrm{Im}\zeta <1$. 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 $\mathrm{Im}\eta >0$ 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 isEvery 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.