I'm currently studying for a complex analysis final and I don't think I've really developed the intuition for conformal mappings yet. I'm attempting a problem from Ahlfors: map the outside of the ellipse $(x/a)^2+(y/b)^2=1$ onto $|w|<1$ with preservation of symmetries. I believe I should use the inverse of the Joukowski transformation at some point (as it maps ellipses to circles) to get a circle of radius $R$ and then rescale. However, I run into trouble when I try to find an $R$ that will work. Any thoughts?
[Math] Conformally mapping an ellipse into the unit circle
analysiscomplex-analysisconformal-geometry
Related Solutions
Suppose your outer circle is the unit circle $C_1=\{z:|z|=1\}$. All you have to do is to get a linear fractional transformation taking $C_1$ to itself and the inner circle $C_2$ to a circle centred at the origin. You might as well rotate $C_2$ so that its centre is on the real axis. Then $C_2$ meets the real axis in points $a$ and $b$ with $-1 < a < b < 1$. The required linear fractional transformation will have real coefficients and take $-1$, $a$, $b$, $1$ to $-1$, $-c$, $c$, $1$ where $0 < c < 1$. The typical linear fractional transformation fixing $-1$ and $1$ has the form $$f_u:z\mapsto\frac{z+u}{uz+1}.$$ We are considering $u$ real and for $u$ to take the unit disc to itself, $|u| < 1$. We need $f_u(a)=-f_u(b)$. This is a quadratic equation in $u$ with two real roots, one of which satsifies $|u| < 1$.
$S(z)=\frac{(z-z_3)(z_2-z_4)}{(z-z_4)(z_2-z_3)}$ is the mobius transformation that maps $z_2\to 1$, $z_3\to 0$ and $z_4\to \infty$.
By orientation principle, a Möbius Transformation(such as $S(z)$) with $S(\Gamma_1)=\Gamma_2$maps the left (or right) hand side of $\Gamma_1$(with respect to ($z_1,z_2,z_3)$) to left (or right) hand side of $\Gamma_2$(with respect to the orientation $(Sz_1,Sz_2,Sz_3)$).
Basically what you want to do is find a map, $T$, which takes $1\to -1$, $i\to 0$ and $-1\to 1$
You can now construct a Möbius Transform $M$ that would take $1\to 1$, $i\to 0$ and $-1\to \infty$
Also, $N$ which would take $-1\to 1$, $0\to 0$ and $1\to \infty$
Notice: Will $T(z)=N^{-1}\circ M(z)$ do the trick?
Best Answer
It is through a special Joukowski transformation $z=\alpha w + \beta/w$ with real constants $\alpha$ and $\beta$ (assuming $z=x+iy$ and $w=u+iv$). The constants are determined using the fact that the boundary of the ellipse is mapped to the boundary of the disk $|w|=1$, or $$ z = x+iy = \alpha (u+iv)+\beta/(u+iv)=\alpha (u+iv)+\beta (u-iv).$$ Then the equation $u^2+v^2=1$ becomes $$ \frac{x^2}{(\alpha+\beta)^2} + \frac{y^2}{(\alpha-\beta)^2}=1$$, from which you can determine $\alpha$ and $\beta$ by $|\alpha+\beta|=a, |\alpha-\beta|=b$.