So I'm getting confused because for $z \in \mathbb{C}$ we have $$\cos(z) = \frac{e^z+e^{-z}}{2} \,\,\, \text{and} \,\,\,\,\ \sin(z) = \frac{e^z-e^{-z}}{2i}$$ where $z = x+iy$ with $x,y \in \mathbb{R}$.

However we know that $\cos(iz) = \cosh(z)$ and $\sin(iz) = i\sinh(z)$ for $z\in\mathbb{C}$ right?

But the definition of the hyperbolic functions is given as $$\cosh(x) = \frac{e^{x}+e^{-x}}{2}\,\,\,\, \text{and} \,\,\,\, \sinh(x) = \frac{e^{x}-e^{-x}}{2i}$$ but this looks exactly the same as the trigonometric functions.

So my question is

Since I couldn't find it on Wikipedia, is $x \in \mathbb{R}$ instead of $x\in\mathbb{C}$ when we are giving the definition of the hyperbolic functions? (sorry for the dummy $x$ in $z = x+iy$ and here)

So fundamentally, can we say that the "only" difference in the definition of the trigonometric functions and the hyperbolic is that the argument (or the exponential) of the hyperbolic functions is real, whereas the argument of the exponentials for the trigonometric ones is complex?


can we say that $\cos(x) = \frac{e^{ix}+e^{-ix}}{2}$ for $x\in \mathbb{R}$, but also $\cos(z) = \cos(x+iy) = \frac{e^{i(x+iy)}+e^{-i(x+iy)}}{2} = \frac{e^{ix-y}+e^{-ix+y}}{2} =\frac{e^{-y}e^{ix}+e^{y}e^{-ix}}{2} $ ?

  • $\begingroup$ The answer to your question is that you don't have the correct definitions of sin and cos. Note the $i$ that should be in the argument. $\endgroup$ – Kaynex Jan 19 '17 at 22:09
  • $\begingroup$ @Kaynex do you mean in the argument of the exponent? Because I do have an $i$ at the exponent, but it is hidden by $z$, indeed I have $e^{x+iy} = e^{x}e^{iy}$ $\endgroup$ – Euler_Salter Jan 19 '17 at 22:14
  • $\begingroup$ How many microseconds did you spend looking for it on Wikipedia? See en.wikipedia.org/wiki/… $\endgroup$ – Rob Arthan Jan 19 '17 at 22:27
  • $\begingroup$ @RobArthan it does never specify if it's real or not though $\endgroup$ – Euler_Salter Jan 19 '17 at 22:29
  • $\begingroup$ Here's a quote from that Wikipedia page: "we can extend the definitions of the hyperbolic functions also to complex arguments. The functions sinh z and cosh z are then holomorphic." How unspecific is that? It also gives the relationships between the hyperbolic and ordinary trigonometric functions. $\endgroup$ – Rob Arthan Jan 19 '17 at 22:31

$$\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}, \sinh(x) = \frac{e^{x} - e^{-x}}{2}$$ $$\cos(z) = \frac{e^{iz} + e^{-iz}}{2}, \cosh(x) = \frac{e^{x} + e^{-x}}{2}$$ Note the $i$ in the argument of the trig functions, these are not in the hyperbolic functions. If you expand each right side out using $e^{iz} = \cos(z) + i\sin(z)$, you'll get the trig function back.

  • $\begingroup$ so is it correct to say that $\cos(x) = \frac{e^{ix}+e^{-ix}}{2}$ for $x\in \mathbb{R}$, but also $\cos(z) = \cos(x+iy) = \frac{e^{i(x+iy)}+e^{-i(x+iy)}}{2} = \frac{e^{ix-y}+e^{-ix+y}}{2} =\frac{e^{-y}e^{ix}+e^{y}e^{-ix}}{2} $ ? $\endgroup$ – Euler_Salter Jan 19 '17 at 22:27
  • $\begingroup$ Yes, it holds for all $z \in \mathbb{C}$ $\endgroup$ – Kaynex Jan 19 '17 at 22:30

For real $x$, Euler’s formula $$ e^{ix}=\cos x+i\sin x $$ together with $e^{-ix}=\cos x-i\sin x$ gives $$ \cos x=\frac{e^{ix}+e^{-ix}}{2} \qquad \sin x=\frac{e^{ix}-e^{-ix}}{2i} $$

If two entire functions (that is, holomorphic over $\mathbb{C}$) coincide over the real line, they are equal, because the set of zeros of their difference has an accumulation point (actually infinitely many).

The power series for the cosine and the sine converge over the whole real line, so the same power series considered for a complex variable converge over $\mathbb{C}$ and define entire functions. Because of the remark above, this is the only possible extension of cosine and sine to entire functions over $\mathbb{C}$. Thus it makes sense to define the cosine and the sine over $\mathbb{C}$ by $$ \cos z=\frac{e^{iz}+e^{-iz}}{2} \qquad \sin z=\frac{e^{iz}-e^{-iz}}{2i} $$ because these are entire functions whose restriction to the real line are the same as the standard cosine and sine.

The same can be said for the hyperbolic cosine and sine: their only sensible extensions to the complex numbers (that is, to entire functions) is $$ \cosh z=\frac{e^{z}+e^{-z}}{2} \qquad \sinh z=\frac{e^{z}-e^{-z}}{2} $$

Note the difference: there is $iz$ in the exponent for the cosine and the sine, besides the $i$ at the denominator for the sine; there is no $i$ for the hyperbolic functions.

If we compute $\cosh(iz)$ and $\sinh(iz)$, we get \begin{align} \cosh(iz)&=\frac{e^{iz}+e^{-iz}}{2}=\cos z\\[6px] \sinh(iz)&=\frac{e^{iz}-e^{-iz}}{2}=i\sin z \end{align} and, similarly, \begin{align} \cos(iz)&=\frac{e^{-z}+e^{z}}{2}=\cosh z \\[6px] \sin(iz)&=\frac{e^{-z}-e^{z}}{2i}=i\sinh z \end{align}

Note also that the standard identities over the real line get preserved over the complex numbers, again because of the remark above. So indeed, for $z=x+iy$, we have $$ \cos z=\cos(x+iy)=\cos x\cos(iy)-\sin x\sin(iy)= \cos x\cosh y-i\sin x\sinh y $$ and $$ \sin z=\sin(x+iy)=\sin x\cos(iy)+\cos x\sin(iy)= \sin x\cosh y+i\cos x\sinh y $$


Please notice that the hyperbolic sine, $\sinh$, is defined as

$$\sinh: \Bbb R \to \Bbb R\\\sinh(x) = \frac{e^x - e^{-x}}{2}$$

whereas the sine for complex numbers is

$$\sin:\Bbb C \to \Bbb C\\ \sin(z) = \frac{e^{iz} - e^{-iz}}{2i}$$

Notice both numerator and denominator are different! And yes, the domain/codomain is also different.

  • $\begingroup$ oh okay my mistake! Another thing though, are the hyperbolic functions defined only as functions from $\mathbb{R}$ to $\mathbb{R}$ ? Can we defined them as complex functions? $\endgroup$ – Euler_Salter Jan 19 '17 at 22:12
  • $\begingroup$ @Euler_Salter they can easily be extended to the complex numbers by keeping the formulas as above, given that we know how to calculate exponentials of complex numbers; I don't know, however, if that will hold any meaning or if there is any other meaningful way of extending $\sinh, \cosh$ to the complex numbers. $\endgroup$ – RGS Jan 19 '17 at 22:17
  • $\begingroup$ No, $\sin z=\dfrac{e^{iz}-e^{-iz}}{2i}$. It's not just the denominator to be different. $\endgroup$ – egreg Jan 19 '17 at 22:41
  • $\begingroup$ @egreg you are right... just today I wrote some annotations on it and already forgot it! thanks for pointing that out! $\endgroup$ – RGS Jan 19 '17 at 22:44

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