Joukowsky Transformation is: $w=z+\frac{1}{z}$ I have been asked about injectiveness of the image with respect to the area bounded inside a circle around the origin with radius $r$ for different values of $r$.
I know that the transformation $\frac{1}{z}$ preserves shapes. And I know that the Joukowsky Transformation transforms circles to ellipses. And for $r=1$ we get a staight line... But I can't figure it out whether it preserves shapes. Is an ellipses can be seen as a circle? And the straight line - as a circle with infinite radius?
Thanks for help...
The Joukowsky transform is injective on every circle of radius $r \neq 1$. However, when $r\neq 1$, $w$ maps the circle of radius $r$ and the circle of radius $1/r$ to the same ellipse, because we can easily check that $w(z) = w(1/z)$. This identity also shows you that $w$ fails to be injective on the unit circle. In particular, the unit circle gets sent to the interval $[-2,2]$, because if $|z|=1$ then $w(z) = z + \overline{z} = 2\text{Re}~z = 2\cos(\arg z)$. Other than these cases, circles of different radii get mapped to different ellipses. So $w$ is injective as a map on $\mathbb{C}\setminus\overline{\mathbb{D}}$ (where $\mathbb{D}$ denotes the unit disc), and also as a map on $\mathbb{D}\setminus \{0\}$, and therefore also conformal onto the image. In both cases the image is $\mathbb{C}\setminus[-2,2]$. If we were to restrict the domain a little more, say $\mathbb{C}\setminus \alpha\overline{\mathbb{D}}$ for $\alpha>1$ or $\beta\overline{\mathbb{D}}\setminus 0$ for $\beta<1$, then $w$ would be a conformal mapping of the domain onto the exterior of an ellipse.