Joukowski map

The holomorphic function $J(w)=-(w+w^{-1})$ maps $D_{+}=\{w:|w|<1,\mathrm{Im}w>0\}$ conformally and bijectively onto the upper half-plane. Its imaginary part is $\mathrm{Im}w(|w{|}^{-2}-1)>0$, and its derivative $w^{-2}-1$ never vanishes there.

Equality of two values implies $(w_1-w_2)(1-1/(w_1{w}_2))=0$; since $|w_1{w}_2|<1$, this proves injectivity. For any target $\zeta$ in the upper half-plane, the quadratic equation $w^2+\zeta w+1=0$ has two roots with product one and neither on the unit circle. Exactly one therefore has modulus below one; the imaginary-part formula forces that root to lie in $D_{+}$, proving surjectivity.

In particular, $z\mapsto e^{iz}$ maps $\{0<\mathrm{Re}z<\pi ,\mathrm{Im}z>0\}$ bijectively to $D_{+}$, and composition gives the conformal map $z\mapsto -2\cos z$ onto the upper half-plane. The two sides of the half-strip map to the outer real rays, while its bottom edge maps to the segment $[-2,2]$.

For $\rho >0$, substitution in the Joukowski map gives the title parametrization. If $\rho \ne 1$, its image is an ellipse with semiaxes $\rho +{\rho }^{-1}$ and $|\rho -{\rho }^{-1}|$ and foci $\pm 2$. Reciprocal radii give the same ellipse in opposite orientations. At $\rho =1$ the image degenerates to $[-2,2]$, each interior point covered once by each unit semicircle.

The Joukowski map restricts to a conformal map from $\{\mathrm{Im}z>0,|z|>1\}$ onto the upper half-plane. Indeed, $\mathrm{Im}g(z)=\mathrm{Im}z(1-|z{|}^{-2})>0$ there. For any $\mathrm{Im}\zeta >0$, the roots of $z^2-\zeta z+1=0$ have product one and cannot lie on the unit circle; exactly one has modulus greater than one, and that root has positive imaginary part. This proves that the map is a bijection.

Its inverse function iswith each factor taken as the principal square root. This product is the holomorphic square root of ${\zeta }^2-4$ that behaves as $\zeta$ at infinity in the upper half-plane. The upper unit semicircle maps to $[-2,2]$; the exterior real rays map to the remaining real rays. The derivative $g^{\prime }=1-z^{-2}$ vanishes only at the boundary endpoints $z=\pm 1$.