0

Find the length of $\{z \in \mathbb C: |z-1|+|z+1|=4\}$.

If we could write $z=x+iy$, then this is an ellipse of the form $$\frac{x^2}{4}+\frac{y^2}{3}=1$$ So $x=2\cos t, y=\sqrt{3}\sin t$, so the path is $$\gamma(t)=2\cos t+i\sqrt{3}\sin t, t \in [0, 2\pi]$$ Thus, the primeter of the ellipse is $$\begin{aligned} \int^{2\pi}_0|\gamma'(t)|dt & =\int^{2\pi}_0|-2\sin t +i\sqrt{3}\cos t|dt\\ & =\int^{2\pi}_0 \sqrt{4\sin^2t+3\cos^2t}dt\\ & =\int^{2\pi}_0\sqrt{3+\sin^2t}dt \end{aligned}$$ Any help with how to go any further? Thanks~

Matata
  • 2,088

1 Answers1

0

In a previous post I give the solution for the arc length of an ellipse in the complex plane, e.g.,

$$ z=a\cos t+ib\sin t,\quad t\in[0,2\pi]\\ s=\int |\dot z|~dt $$

The solution can be expressed in terms of the comple elliptic integral of the second kind or the Gauss hypergeometric function as follows:

$$ \begin{align} s &=\int_0^{2\pi}b\sqrt{A\sin^2\theta+1}~d\theta,\quad A=\left(\frac{a^2-b^2}{b^2}\right)\\ \\ &=4b\text{E}(-A)=4b\frac{\pi}{2}~ _2\text{F}_1\left(-\frac 12,\frac 12;1;-A \right) \end{align} $$

For $a=2$ and $b=\sqrt{3}$, we find that $s\approx11.7397$.

Cye Waldman
  • 7,524