Exponential of 2 m i Arccotangent of p

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Theorem

$\map \exp {2 m i \arccot p} \paren {\dfrac {p i + 1} {p i - 1} }^m = 1$


Proof

Let $z = \arccot p$.

Then:

\(\ds p\) \(=\) \(\ds \frac {\cos z} {\sin z}\) Definition of Real Arccotangent
\(\ds \leadsto \ \ \) \(\ds p\) \(=\) \(\ds i \dfrac {\map \exp {i z} + \map \exp {-i z} } {\map \exp {i z} - \map \exp {-i z} }\) Euler's Cotangent Identity
\(\ds \) \(=\) \(\ds \frac {i \paren {\map \exp {2 i z} + 1} } {\map \exp {2 i z} - 1}\)
\(\ds \leadsto \ \ \) \(\ds p \paren {\map \exp {2 i z} - 1}\) \(=\) \(\ds i \paren {\map \exp {2 i z} + 1}\)
\(\ds \leadsto \ \ \) \(\ds \paren {p - i} \map \exp {2 i z}\) \(=\) \(\ds p + i\)
\(\ds \leadsto \ \ \) \(\ds \map \exp {2 i z}\) \(=\) \(\ds \frac {p + i} {p - i}\)


So we have:

\(\ds \map \exp {2 m i \arccot p}\) \(=\) \(\ds \paren {\map \exp {2 i \arccot p} }^m\)
\(\ds \) \(=\) \(\ds \paren {\frac {p + i} {p - i} }^m\)


Now:

\(\ds \paren {\frac {p + i} {p - i} } \paren {\frac {p i + 1} {p i - 1} }\) \(=\) \(\ds \frac {p^2 i + p i^2 + p + i} {p^2 i - p i^2 - p + i}\)
\(\ds \) \(=\) \(\ds \frac {p^2 i + i} {p^2 i + i}\)
\(\ds \) \(=\) \(\ds 1\)

and the result follows.

$\blacksquare$


Sources

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