# $|\cosh z|^2 = \cosh^2x + \sin^2y$

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?

• First, let $z=$\pi i/2$, and calculate your result, and the one you're trying to get, and see which one is right. – Gerry Myerson Feb 26 '19 at 9:18 • Alternatively, replace all the trig and hyperbolic functions with exponentials (e.g.,$\cos w=(1/2)(e^{iw}+e^{-iw})$) and see what falls out. – Gerry Myerson Feb 26 '19 at 9:20 • Your$-$version is what I got, too. Where did you read the incorrect$+\$ version? – J.G. Feb 26 '19 at 9:49
• @J.G. Arfken (it's translated so it's probably a tipo) – Avesta Sabayemoghadam Feb 26 '19 at 9:56
• Ah, that book, right. – J.G. Feb 26 '19 at 10:05

$$|\cosh z|^2 = \cosh^2x + \sin^2y$$ 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.

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.

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

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