1
$\begingroup$

The Joukowski map $J : HD \rightarrow H$ is defined by $$f(z) = -\frac{1}{2}(z + z^{-1})$$ where $HD = \{x+iy : |x+iy| < 1, y > 0\}, H = \{x+iy : y > 0\}.$

It can be shown that $f$ is a conformal mapping from $HD$ surjectively to $H$. So it maps the points inside the upper half of unit circle to the upper half plane.

Fix $\theta \in (0, \pi)$, then the point a the radius of unit circle can be described as $z = re^{i\theta},$ $r \in (0,1)$.

Geometrically, what does the map $f$ do to this line ? It stretch the line by mapping a point near $0$ to $\infty$ and a point on the radius with $|z| \rightarrow 1$ to where ?

Is there a way to see what it does to the boundary of $HD$ ? like what is the geometric image of $f(L)$ where $L$ is a line from $-1$ to $1$, or $f(Cr)$ where $Cr$ is an arc of unit circle ranging from angle $0$ to $\pi$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.