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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?

Neil Iyer
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  • 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 08 '22 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 --> https://math.stackexchange.com/q/3729184/399263 – zwim Apr 08 '22 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... – DonAntonio Apr 08 '22 at 19:47
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    @DonAntonio since when it is not true for $x\in\mathbb C$ ? – zwim Apr 08 '22 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. – DonAntonio Apr 08 '22 at 19:51

2 Answers2

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Use the definition of the trigonometric complex function:

$$\cos z=\frac{e^{iz}+e^{-iz}}2\implies \cos i=\frac{e^{i\cdot i}+e^{-i\cdot i}}2=\frac{e^{-1}+e^{1}}2=\cosh1$$

For the sine you have

$$\sin z=\frac{e^{iz}-e^{-iz}}{2i}$$

DonAntonio
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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.

Joseph DeGaetani
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