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So I have the following computation $$ e^{\pi i(\beta+\beta')}(e^{2\pi i(\alpha+\gamma)}-e^{-2\pi i \beta'}), $$

everything is a constant. For a bit of context this is a computation appearing in finding the monodromy of the hipergeometric differential equation.

And trying to simplify leaves me stuck at

$$ e^{\pi i(\beta+\beta')}(e^{2\pi i(\alpha+\gamma)}-e^{-2\pi i \beta'})=e^{\pi i(\alpha+\beta+\gamma)}e^{\pi i(\alpha+\beta'+\gamma)}-e^{\pi i(\beta-\beta')}. $$

I need to arrive at something like

$$ 2\pi ie^{\pi i(\alpha+\beta+\gamma)}\sin\pi(\alpha+\beta'+\gamma). $$

Any hints or suggestions?

Anne Bauval
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Celepharn
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1 Answers1

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$$e^{\pi i(\beta+\beta')}(e^{2\pi i(\alpha+\gamma)}-e^{-2\pi i \beta'})$$ $$=e^{i\pi(\alpha+\beta+\gamma)}(e^{i\pi(\alpha+\gamma+\beta')}-e^{-i\pi(\alpha+\gamma+\beta')})$$ $$=e^{i\pi(\alpha+\beta+\gamma)}2i\sin\pi(\alpha+\gamma+\beta').$$

More generally, $$e^{ix}-e^{iy}=e^{i(x+y)/2}(e^{i(x-y)/2}-e^{-i(x-y)/2})=e^{i(x+y)/2}2i\sin((x-y)/2).$$

Anne Bauval
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    Indeed. It might be hard for a novice to "guess" this sort of balanced adjustment to find a sine-function (or cosine, with a plus...), without knowing the trick already. :) – paul garrett Jul 30 '24 at 21:54