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I'm trying to prove : $|\cosh z|^2 = \cosh^2x + \sin^2y$ I know that: $|\cos z|^2 = \cos^2x + \sinh^2y$ My procedure is: $|\cosh z|^2 = |\cos iz|^2 = \cos^2y + \sinh^2x$ And since: $\cos^2x + \sin^2x = 1$ $\cosh^2x - \sinh^2x = 1$ So I get: $|\cosh z|^2 = \cosh^2x - \sin^2y$

What is wrong with my procedure?

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  • $\begingroup$ First, let $z=$\pi i/2$, and calculate your result, and the one you're trying to get, and see which one is right. $\endgroup$ – Gerry Myerson Feb 26 '19 at 9:18
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    $\begingroup$ Alternatively, replace all the trig and hyperbolic functions with exponentials (e.g., $\cos w=(1/2)(e^{iw}+e^{-iw})$) and see what falls out. $\endgroup$ – Gerry Myerson Feb 26 '19 at 9:20
  • $\begingroup$ Your $-$ version is what I got, too. Where did you read the incorrect $+$ version? $\endgroup$ – J.G. Feb 26 '19 at 9:49
  • $\begingroup$ @J.G. Arfken (it's translated so it's probably a tipo) $\endgroup$ – Avesta Sabayemoghadam Feb 26 '19 at 9:56
  • $\begingroup$ Ah, that book, right. $\endgroup$ – J.G. Feb 26 '19 at 10:05
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$|\cosh z|^2 = \cosh^2x + \sin^2y$ can't be right, as this would mean $|\cosh z|^2 \geq 1$ for all $z \in \mathbb{C}$, which is impossible by Liouville. What you computed seems right, though.

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Note that $\left\lvert\cosh\left(\frac{\pi i}2\right)\right\rvert^2=0$. Therefore, the equality that you are trying to prove is false.

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I give a comprehensive derivation because I noticed that the formula in the heading contains a sign error. In the meantime this error has been corrected in the text of the OP.

Let $z=x + i y$ with $x$ and $y$ real.

Then we have

$$f=\cosh(z) = \cosh(x + i y) = \cosh (x) \cosh(i y) + \sinh(x) \sinh(i y)$$

Since $\cosh(i y) = \cos(y)$, $\sinh(i y) = i \sin(y)$ we can proceed

$$f = \cosh (x) \cos( y) + i \sinh(x) \sin(y)$$

Now since $x$ and $y$ are assumed to be real we have

$$|f|^2 = \Re(f)^2+\Im(f)^2 = \cosh (x)^2 \cos( y)^2+\sinh(x)^2 \sin(y)^2\\= \cosh (x)^2(1- \sin( y)^2)+\sinh(x)^2 \sin(y)^2 \\ =\cosh (x)^2 - \sin( y)^2(\cosh (x)^2-\sinh(x)^2)$$

Since $(\cosh (x)^2-\sinh(x)^2)= 1$ we get finally

$$|\cosh(x + i y)|^2= \cosh (x)^2 -\sin( y)^2 $$

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