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In my text book I saw that ${|\cosh z|}^2={\cos}^2x+{\sinh}^2y$

But when I tried deriving it myself I got this: $${|\cosh z|}^2={\cos}^2y+{\sinh}^2x$$

See my working below:

$$\cosh z=\frac{1}{2}(e^z+e^{-z})=\frac{1}{2}(e^{x+yi}+e^{-x-yi})=\frac{1}{2}(e^x\cos y+ie^x\sin y+e^{-x}\cos y-ie^{-x}\sin y)$$

Solving further and collecting like terms gives:

$$\frac{1}{2}(e^x+e^{-x})\cos y + \frac{1}{2}(e^x-e^{-x})i\sin y=\cosh x\cos y+i\sinh x\sin y$$

Therefore we can get the square of the absolute value of $\cosh z$

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

Am I wrong or is my textbook wrong? Or are we both right?!

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    $\begingroup$ You seem to be correct to me. $\endgroup$ – Patrick Stevens Aug 30 '15 at 22:13
  • $\begingroup$ @PatrickStevens but is the textbook right as well? $\endgroup$ – Obinoscopy Aug 30 '15 at 22:16
  • $\begingroup$ @Obinoscopy : Both cannot be right. One expression is bounded for all real $x$ and the other is bounded for all real $y$. It cannot be bounded in both. $\endgroup$ – DisintegratingByParts Sep 1 '15 at 0:27
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The identity you quote as being the textbook's answer - $|\cosh(x+i y)|^2 = \cos(x)^2 + \sinh(y)^2$ - is false in the case $y=0$ for all $x \not = 0$.

Your identity and proof are correct. It is the case that $|\cosh(x+i y)|^2 = \sinh(x)^2 + \cos(y)^2$.

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For real $z$, say, $z := x + 0 i$, we have $|\cosh z|^2 = \cosh^2 x$, which is unbounded, but substituting this real quantity in your book's formula gives $\cos^2 x$, which is bounded, and so the book's formula is wrong. Your derivation gives the correct formula.

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