Maps circles onto ellipses
Show that the mapping $w=z+\frac1z$ maps circles $|z|=p(p\ne 1)$ onto ellipses $$\frac{u^2}{(p+\frac1p)^2}+\frac{v^2}{(p-\frac1p)^2}=1.$$
I can parametrize the circle by $z(t)=pe^{it}, \ 0 \leq t\leq 2\pi$. Then, $w(t)=pe^{it}+\frac{1}{pe^{it}},$ but then how can I show that this does map circles onto ellipses?
$\endgroup$2 Answers
$\begingroup$An ellipse of the type $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$can be parametrized by $x=a\cos[\theta], y=b\sin[\theta]$. So in our case we want $x=(p+p^{-1})\cos[\theta]$, $y=(p-p^{-1})\sin[\theta]$. Notice that $x,y$ are, respectively, the real and imaginary parts of $w$. But we know $w=p(\cos[\theta]+i\sin[\theta])+\frac{1}{p}(\cos[\theta]-i\sin[\theta])$, then the results are immediate. Notice that you made an arithmetic mistake in computing $w$, which may be why.
The geometrical picture is the map "dilates" the unit region by a stretch on $x,y$ coordinates, which is not intuitively clear from the formula given. But it should be clear by now.
$\endgroup$ 6 $\begingroup$The so called Joukowski function$$J(z):={1\over2}\left(z+{1\over z}\right)$$ is a familiar animal in complex analysis courses. The circle $$\gamma_r:\quad t\mapsto r e^{it}\qquad(0\leq t\leq 2\pi)$$ is transformed by $J$ into the curve $$ \eqalign{J(\gamma_r):\quad t\mapsto w(t)&={1\over2}\left(r e^{it}+{1\over r}e^{-it}\right)\cr &={1\over 2}\left(r+{1\over r}\right)\cos t +i \>{1\over 2}\left(r-{1\over r}\right)\sin t\qquad(0\leq t\leq 2\pi)\ .\cr}$$ The image curve is obviously an ellipse with semiaxes $$a:={1\over2}\left(r+{1\over r}\right),\qquad b={1\over2}\left|\,r-{1\over r}\right|\ .$$ It is encircling the interval $[-1,1]$ counterclockwise when $r>1$, and clockwise when $r<1$. All these ellipses are confocal with foci $\pm1$.
$\endgroup$
