The function $f(z)=z+\frac{1}{z}$ maps $\Omega$ biholomorphically on the upper plane

Solution 1:

If $z_1\ne z_2$ and $|z_1|, |z_2|>1$, then $$\left|\frac{1}{z_2}-\frac{1}{z_1}\right| = \frac{|z_1-z_2|}{|z_1||z_2|}<|z_1-z_2| \tag1$$ hence $$\frac{1}{z_2}-\frac{1}{z_1}\ne z_1-z_2 \tag2$$ This gives injectivity.

Using polar coordinates $z=re^{it}$, we find that $\operatorname{Im}(1/z)=-|z|^{-2}\operatorname{Im}z = -r^{-1}\sin t$. So, the curve $\operatorname{Im}(z+1/z)=a$ can be written as $$(r-r^{-1})\sin t=a$$ which can be solved for $r$ or for $t$. Either way, you will see that along the curve $$\operatorname{Re}(z+1/z) =(r+r^{-1})\cos t$$ takes on all real values, being continuous and unbounded in both directions. This implies surjectivity as a by-product.