Joukowski mapping a circle onto an ellipse problem The Joukowski mapping is defined by
$$\displaystyle w = J(z) = \frac{1}{2} \left(z + \frac{1}{z}\right)$$
where $ z = x + yi $

Show that
$J$ maps the circle $|z| = r (r > 0, r \neq 1)$ onto the ellipse
$$ \frac{u^2}{\left[\frac{1}{2}\left(r + \frac{1}{r}\right)\right]^2} +\frac{v^2}{\left[\frac{1}{2}\left(r - \frac{1}{r}\right)\right]^2} = 1 $$
which has a foci at $\pm 1$.

Would I have make use of the fact that $z = re^{i\theta} = r(\cos \theta + i\sin \theta)$ and $ u + iv = w = J(z) = \frac{1}{z} \left(z + \frac{1}{z}\right) = \frac{1}{z} \left(re^{i\theta} + \frac{1}{r} e^{-i\theta}\right)$?

HINT...You are almost there: you have $$u+iv=\frac 12\left(r(\cos\theta+i\sin\theta)+\frac{1}{r(\cos\theta+i\sin\theta)}\right)$$
$$=\frac 12\left(r(\cos\theta+i\sin\theta)+\frac 1r(\cos\theta-i\sin\theta)\right)$$
So now you can get $\cos\theta$ and $\sin\theta$ in terms of $u$ and $v$ and use the Pythagorean identity to eliminate $\theta$
