Find the analytic function $w(z)$ that performs a conformal mapping of the region $$\Omega \equiv \left \{ |z|>1,|z-1|<1 \right \}$$ to the upper half-plane $\text{Im}(w)>0$
My attempt:
First, I entered a new variable $$w_1=\frac{z-1}{z+1}$$ The mapping $z\mapsto w_1$ transfers the circle $|z|=1$ to the real axis, and the line $z=1$ to the point $w_1=0$. Under that mapping, the region $\Omega$ is mapped into the band $-\frac12<\text{Re}(w_1)<\frac12$, $|\text{Im}(w_1)|>1$
After that I used the Schwarz-Christian transformation, which maps the strip to the upper half-plane $$w_2=\exp\left[\pi i \cdot \frac{w_1+\frac12}{w_1-\frac12}\right]$$
And it turns out $$w(z)=w_2(w_1(z))=\exp\left[\pi i \cdot \frac{\frac{z-1}{z+1}+\frac12}{\frac{z-1}{z+1}-\frac12}\right]$$
Am I reasoning correctly in solving this problem? Am I solving it correctly? I would be very glad if you could tell me