Let \(D\) be the region obtained by deleting the real interval \([0, 1)\) from \(\DD\); find a conformal map from \(D\) to \(\DD\).
Solution
On \[ D=\{0<|z|<1:\arg z\in(0,2\pi)\} \] choose the square-root branch \[ s(z)=\sqrt z=|z|^{1/2}e^{i\arg z/2}, \qquad 0<\arg z<2\pi. \] Then \(s\) maps \(D\) biholomorphically onto the upper half-disk \[ U=\{w:|w|<1,\ \Im w>0\}. \]
Next \[ T(w)=\frac{1+w}{1-w} \] maps \(U\) biholomorphically onto the first quadrant: the diameter \((-1,1)\) goes to the positive real axis, while the upper semicircle goes to the positive imaginary axis. Therefore \(T(w)^2\) maps \(U\) biholomorphically onto the upper half-plane. Finally the Cayley map \[ C(\zeta)=\frac{\zeta-i}{\zeta+i} \] maps the upper half-plane onto \(\mathbb D\). Thus one required map is \[ \boxed{ F(z)= \frac{\left(\frac{1+\sqrt z}{1-\sqrt z}\right)^2-i} {\left(\frac{1+\sqrt z}{1-\sqrt z}\right)^2+i}, } \] with the square-root branch specified above.