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http://www.webassign.net/zillengmath4/20.2.pdf p.2.

The conformal map $z+\frac{1}{z}$ maps circles $|z|=r$ to ellipses and $arg(z)=\theta$ to hyperbolas.

I believe one can display both using the same equations, but I have only managed to display the ellipse and cannot understand the hyperbola.

Basically some sources (and I) claim that

$$\left(r+\frac{1}{r}\right)\cos\theta+i\left(r-\frac{1}{r}\right)\sin \theta$$

is an equation of an ellipse, when $r$ is fixed and $\theta$ is varied. I think this is fairly clear.

But of a hyperbola when $\theta$ is fixed and $r$ is varied. Why?

Hagen von Eitzen
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mavavilj
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2 Answers2

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Note that the product of $u=(r+\frac1r)+(r-\frac1r)$ and $v=(r+\frac1r)-(r-\frac1r)$ is constant, just as we know it for $y\cdot x$ in the hyperbola $y=\frac 1x$. These $u$ and $v$ occur (up to sign?) as real and imaginary part if you first "rotate away" the $\cos \theta+i\sin \theta$ and then multiply by $1+i$ (i.e., (stretch-)rotate by $45$ degrees)

Hagen von Eitzen
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Consider ordered pairs parametrized by $$(x(r), y(r)) = ((r + r^{-1})\cos \theta, (r - r^{-1})\sin \theta), \quad r \ne 0, \quad \theta \not\in\{k\pi/2\}_{k\in \mathbb Z}.$$ Then $$((x \sin \theta)^2, (y \cos \theta)^2) = ((r^2 + r^{-2} + 2)(\cos \theta \sin \theta)^2, (r^2 + r^{-2} - 2)(\cos \theta \sin \theta)^2)$$ and it now becomes clear that $$(x \sin \theta)^2 - (y \cos \theta)^2 = 4(\cos \theta \sin \theta)^2,$$ or in standard form, $$\frac{x^2}{4 \cos^2 \theta} - \frac{y^2}{4 \sin^2 \theta} = 1.$$ This demonstrates that the parametrization is some subset of a hyperbola for constant $\theta$.

heropup
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  • So the hyperbolicity is merely seen by squaring the thing that one got for the ellipses. How is this related to $arg(z)=\theta$ though. Since the question asks whether $arg(z)=\theta$ is mapped to hyperbolas. – mavavilj Feb 04 '16 at 18:41
  • @mavavilj I edited my post to make the relationship explicit, as it appeared that my previous hint was not sufficient. – heropup Feb 04 '16 at 18:46
  • What does "subset of a hyperbola" mean? How does the squaring process "reduce" back to where we started? – mavavilj Feb 05 '16 at 01:47