MHB Prove sinh(z) Identity: Show |sinh(z)|^2 = sin^2x + sinh^2y

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The discussion centers on proving the identity |sinh(z)|^2 = sin^2x + sinh^2y. A participant found that using the identities cos(iy) = cosh(y) and sin(iy) = i sinh(y) simplifies the proof. Another user pointed out a potential typo in the original equation, suggesting it should be |sinh(z)|^2 = sinh^2x + sin^2y for accuracy. The original poster indicated they would mark the question as solved after their findings. The conversation highlights the importance of precise notation in mathematical identities.
ognik
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Show $ |sinh(z)|^2 = sin^2x + sinh^2y $
Since I posted this, I found new info - cos(iy) = cosh(y) and sin(iy) = i sinh(y) which made the above easy; don't want to bother anyone so will mark this solved.
 
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ognik said:
Show $ |sinh(z)|^2 = sin^2x + sinh^2y $
Since I posted this, I found new info - cos(iy) = cosh(y) and sin(iy) = i sinh(y) which made the above easy; don't want to bother anyone so will mark this solved.

Hey ognik,

Just for the record, it appears there is a typo in there.
It should be for instance
$$ |\sinh(z)|^2 = \sinh^2x + \sin^2y $$
otherwise it's not true. :eek:
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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