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Multiplying the identity $$ \cosh^2 x - \sinh^2x = 1 $$ with $c^2$ and defining $a^2= c^2\cosh^2$ and $b^2= c^2\sinh^2$ we get a hyperbolic version of the pythagorean theorem: $$ a^2-b^2 = c^2 $$ In analogy to this question I try to find a geometric interpretation by factorizing this relation $$ (a + jb)(a - jb) = c^2 $$ where I used the imaginary number $j^2 =1$ so I get a product of two Split-complex numbers. What I find confusing is that I could also factorize this relation using real numbers $$ (a + b)(a - b) = c^2 $$ So there are two ways to factorize the hyperbolic pythagorean theorem.

Now my question is: How does one interpret the hyperbolic pythagorean theorem geometrically? And can one interpret the two ways of factorization also geometrically in two different ways?

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    imaginary number $j^2=1$?? For the imaginary unit $i^2=-1$ one has $(a+ib)(a-ib)=a^2+b^2$. – 2017-02-21
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    Note that this is not the hyperbolic pythagorean theorem, the correct such version is $\cosh c=\cosh a\cosh b$. – 2017-02-21
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    @PeterMelech OP mentions that $j$ is the imaginary unit in the *split-*complex numbers, not the usual complex numbers. – 2017-02-21
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    @Travis Thanks! Actually I never heard of these numbers. That´s why I was confused! – 2017-02-21
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    @ReneSchipperus As mentioned at https://en.wikipedia.org/wiki/Hyperbolic_function#Useful_relations the relation I start from is "similar to the Pythagorean trigonometric identity". – 2017-02-21
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    @PeterMelech They're probably not described much in isolation in part because, (1) while they admit a natural real quadratic form $a + jb \to a^2 - b^2$, this form is indefinite (but still nondegenerate), (2) as an $\Bbb R$-module, the split-complex numbers is isomorphic to the more familiar $\Bbb R \oplus \Bbb R$. – 2017-02-21
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    @Travis Yes, but the connection to the Minkowski plane is definitely interesting. Thanks. – 2017-02-21

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