# show that Joukowski transform is one-to-one in the upper half outside the unit disk

I have a problem on showing how the Joukowski transform $w=J(z) = .5(z + 1/z)$ takes the upper half plane, $|z| \gt 1$, one-to-one into the w upper half plane. I have shown how the unit disk itself collapses onto the real axis and how points outside it approach $(x,0)$ as $|z| \rightarrow 1$ from above. But I think I am supposed to show that there is a single-valued inverse function, and this is what I'm stuck on. Solving $J(z)=w$ for $z = w±\sqrt{w^2-1}$ and I have a rather larger formula for w in terms of $z = x + iy$, which seems like maybe more what I should be concerned with since after all the other condition is that $Im(z) \gt 0$, but also maybe I should be looking at the branch cuts except I'm not quite sure how to deal with that either. Any thoughts? Thank you.

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One way to prove injectivity is to notice that for any distinct $z_1,z_2$ in your domain $$\left|\frac{1}{z_1}-\frac{1}{z_2}\right| = \frac{|z_1-z_2|}{|z_1|\,|z_2|}<|z_1-z_2| \tag1$$ hence $$\frac{1}{z_1}-\frac{1}{z_2}\ne z_1-z_2 \tag2$$ which is equivalent to $$z_1+\frac{1}{z_1}\ne z_2+\frac{1}{z_2} \tag3$$