# How to find the sine of an imaginary number? [duplicate]

So I was trying to prove to myself that $$i^i$$ is equal to a real number. By doing that I encountered a problem, how can you find the sine or cosine of an imaginary number.

So let me show you my math:

$$e^{ix} = \cos(x) + i\sin(x)$$

$$e^{i(i)} = \cos(i) + i\sin(i)$$

$$e^{-1} = \cos(i) + i\sin(i)$$

To summarize, I get that $$e^{-1}$$ is a real number, but how is $$\cos(i) + i\sin(i)$$ one and how do you calculate it?

• use hyperbolic functions $\sinh(x)=\frac 12(e^x-e^{-x})$ and $\cosh(x)=\frac 12(e^x+e^{-x})$ . Express $\sin(ix)$ in function of $\sinh(x)$.
– zwim
Apr 8, 2022 at 19:42
• Look also at my answer here to calculate $z^u$ for any complexes, and some applications in the linked posts to it --> math.stackexchange.com/q/3729184/399263
– zwim
Apr 8, 2022 at 19:44
• Your first formula (which is Euler's Formula) is true iff $\;x\in\Bbb R\;$, and thus your second formula is wrong... Apr 8, 2022 at 19:47
• @DonAntonio since when it is not true for $x\in\mathbb C$ ?
– zwim
Apr 8, 2022 at 19:49
• @zwim thought the OP meant $\;x\;$ is a real number and is thus using Euler's Formula. This is usually done that way in order to avoid a circular definition. Apr 8, 2022 at 19:51

$$\cos z=\frac{e^{iz}+e^{-iz}}2\implies \cos i=\frac{e^{i\cdot i}+e^{-i\cdot i}}2=\frac{e^{-1}+e^{1}}2=\cosh1$$
$$\sin z=\frac{e^{iz}-e^{-iz}}{2i}$$
One option is to do as DonAntonio did, another option is as follows. Given $$e^{ix}=\cos(x)+i\sin(x)$$ let $$x=\frac{\pi}{2}$$. Then we see that $$e^{i\frac{\pi}{2}}=i.$$ Raising both sides to the $$i^{th}$$ power gives $$e^{i^2\frac{\pi}{2}}=e^{\frac{-\pi}{2}}=i^i.$$ Since the former is a real number, so too is the latter. We sweep aside issues where things can take on multiple values, with the promise that taking other representatives also return real numbers.