I'm trying to prove that

$|\cot (x+iy)|^2=\frac{\cos^2x+\sinh^2y}{\sin^2x+\sinh^2y}$

for $x+iy \in \mathbb{C} \setminus \pi\mathbb{Z}$.

I've tried to use the identities $\cos(iy)=\cosh y$ and $\sin(iy)=i\sinh y$, but I obtain

$|\cot (x+iy)|^2=\frac{\cos^2(x) \cosh^2 y+\sin^2(x) \sinh^2y}{\sin^2(x)\cosh^2 y+\cos^2(x)\sinh^2y}$

What did I miss?


You have missed nothing except simplification

Using $\cos^2x+\sin^2x=\cosh^2y-\sinh^2y=1$, eliminate $\cosh y,\sin x$ from the numerator and $\cosh y,\cos x$ from the denominator.



Again, $\cos^2x+\sinh^2y=1-\sin^2x+\cosh^2y-1=\cosh^2y-\sin^2x$

and $\sin^2x+\sinh^2y=1-\cos^2x+\cosh^2y-1=\cosh^2y-\cos^2x$(this is another simplified form)

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  • $\begingroup$ And $\cos^2(x)\cosh^2y+\sin^2(x)\sinh^2y=\cos^2x(1+\sinh^2y)+(1-\cos^2x)\sinh^2y$ thanks! $\endgroup$ – Fuente Oct 24 '12 at 10:46
  • $\begingroup$ @Fuente, $\cosh^2y=1+\sinh^2y$, please find the updated answer. $\endgroup$ – lab bhattacharjee Oct 24 '12 at 10:48
  • $\begingroup$ I know, it was a typo .) $\endgroup$ – Fuente Oct 24 '12 at 10:50

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