What are the circular-to-hyperbolic trig identities (eg, $\cos(ix)=\cosh(x)$) trying to tell me?

Question

I have been wondering about the utility of

$$\begin{align} \cos(ix)&=\cosh (x) \\ \sin(ix)&=i\sinh(x) \\ \cosh(ix)&=\cos(x) \\ \sinh(ix)&=i\sin(x) \end{align}$$

I feel like these are telling me something profound about a bridge between the real and imaginary numbers and the real/imaginary exponential function.

How can each of these functions take imaginary arguments that then translate back into real arguments of a different function. Whilst I am aware of plugging these arguments into the exponential definitions this seems just like "it works so it;s true" solution. I guess I am struggling to grasp why taking the trig functions into complex numbers can be the same as the hyperbolics in the reals and vice versa? Is it something to do with them being conic sections?

Does this also allow greater flexibility in flicking between complex and real domains have useful application?

A site search for "trigonometric hyperbolic functions" yields over 300 results, many of which look relevant. "Unifying the connections between the trigonometric and hyperbolic functions" (currently the 6th item listed by relevance) has a nice survey of results in the question itself; the answers offer further insights. If these aren't satisfactory, please try to articulate what you find lacking about them, so that people don't spend time unhelpfully re-articulating them — Blue, Apr 12 '21 at 09:50
Well it tells you that $,e^z=\cosh z+\sinh z,$ and $,e^{ix}=\cos x+i\sin x,$ are the same thing if $z=ix$. — Raymond Manzoni, Apr 12 '21 at 10:03
Yes I have read the unifying post but found the leading answer a little lacking. For instance it ends with saying "remember real <-> imaginary" without any real context of what that means. I shall edit my post to clarify my question — Jon Percival, Apr 12 '21 at 10:03

score 2 · Answer 1 · answered Apr 12 '21 at 09:52

2

For example let $f(z)= \cos z$ for $z \in \mathbb C.$

We all know that $|f(x)| \le 1$ for real $x$. Hence $f$ is bounded on $ \mathbb R.$

But

$$f(ix)= \cosh x$$

for $x \in \mathbb R.$ This shows that $f$ is unbounded on $ \mathbb C.$

answered Apr 12 '21 at 09:52

Fred

78,422

Great. Thank you for that insight. So being able to flick between real and imaginary parts in these functions can give aid in the analysis of these functions in these different domains – Jon Percival Apr 12 '21 at 10:11

What are the circular-to-hyperbolic trig identities (eg, $\cos(ix)=\cosh(x)$) trying to tell me?

1 Answers1