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I can't find any explanations online as to why the following holds:

$$(-1)^{-i} = e^\pi$$

I assume there's a simple explanation pursuant to the rules of complex number manipulation, but is there any further insight on a deeper or maybe geometric level?

I'm curious how the $i$ on the left, a complex expression, disappears and you get a real number answer on the right.

Thanks

Edit: I'm getting good answers below, but the geometric interpretation is still elusive to me... I thought raising to the power of i has the effect of rotating the initial input 90 degrees in the complex plane; so I'm not clear how these expressions apparently get us totally out of the complex plane and just down purely on the real line. Forgive me if this is way off. thanks.

Svenn
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    Although we have to be careful with manipulations , it might be useful that one branch gives $i^i=e^{-\pi/2}$ – Peter Jul 17 '24 at 19:27
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    Since you ask for geometric insight: Ignorning branches: Consider $z_\theta=e^{i\theta}$. When $\theta=0$ then $z_0=1$, an arrow if you will pointing from the origin to $(1,0)$. Now let's rotate $z_0$, i.e. the arrow $(1,0)$, by a full $\theta=\pi$ anti-clockwise to get $z_\pi=e^{i\pi}=-1$, an arrow pointing to $(-1,0)$. You can imagine all the other possibilities inbetween. So if you believe $e^{i\pi}=-1$, then plugging into your equation, $(e^{i\pi})^{-i}=e^{-i^2\pi}=e^{\pi}$ as expected. By the way, all real numbers ar IN the complex plane, but not all complex numbers are real. – pshmath0 Jul 17 '24 at 20:09
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    Note you shouldn't be surprised to have an expression on one side containing complex numbers equal to an expression on the other side that is a real number. After all, you aren't surprised by the equations $i^2= -1$ or $(1+i)^8 = 16$ or $(1+i)(1-i)=2$, right? – MPW Jul 17 '24 at 20:14
  • You are both correct, I should have remembered that. As you may have guessed, I find $^$ exciting because of the 19.9990999 phenomenon. – Svenn Jul 17 '24 at 20:40

2 Answers2

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By arbitrary convention, $~-1~$ is often set to $~e^{i\pi},~$ rather than $~e^{i(\pi + 2k\pi)} ~: ~k \in \Bbb{Z}.~$

Then

$$(-1)^{-i} = \left[ ~e^{i\pi} ~\right]^{-i} = e^{-(i^2)\pi} = e^\pi.$$

The comment of Mark Viola, following this answer, brings up a point, which I will repeat below. Here, I am getting into an area which somewhat exceeds my formal education of chapter 1 of "An Introduction to Complex Function Theory" (Palka).

However, the comment of Mark Viola does make sense to me, and so bears repeating:

Excerpting Mark Viola's comment:

...The correct approach (for this answer) is to use the definition

$\displaystyle z^C = e^{C\log(z)},~$

where $~\log(z)~$ is multivalued.

Then, we see that

$\displaystyle (−1)^{−i} = e^{−i\log(−1)} = e^{−i(i(2n+1)\pi)} ~: ~n \in \Bbb{Z} = e^{(2n+1)\pi} ~$

which is multivalued.

This is consistent with the (somewhat informal) result that I gave at the start of my answer.

Then, following the (arbitrary) convention that the principal log of $~(-1)~$ is $~i\pi,~$

you have that

$$e^{−i \times [ ~\text{principal} ~\log(−1) ~]} = e^{-i(i\pi)} = e^{\pi}.$$

user2661923
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    You are using the property $(z^a)^b = z^{ab}$ which is in general false for complex numbers, so some extra justification is needed here. – jjagmath Jul 17 '24 at 19:38
  • @jjagmath +1 : I agree with your comment. Answer edited. – user2661923 Jul 17 '24 at 19:43
  • This answer relies on a generally false premise. The correct approach is to use the definition $z^C=e^{C\log(z)}$, where $\log(z)$ is multivalued. Then, we see that $$(-1)^{-i}=e^{-i\log(-1)}=e^{-i(i(2n+1)\pi)}=e^{(2n+1)\pi}$$which is multivalued. Please delete this post or correct it. – Mark Viola Jul 17 '24 at 21:15
  • @MarkViola I have just updated the end of my answer to share the idea in your comment. As I state in my answer, while your idea somewhat exceeds my formal education in this area, your idea still makes sense to me. Please advise if you think that a further editing is warranted. – user2661923 Jul 18 '24 at 00:15
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See my answer to this question.

Complex exponential expressions are generally multivalued. This comes from the periodicity of the exponential function. For nonzero $z$, one defines $z^w$ as $e^{w\log z}$; there are countably infinitely many values for $\log z$, all differing by integral multiples of $2\pi i$. This means there are multiple values of $z^w$, unless $w$ happens to be an integer , in which case there is only one (because then $2\pi w i$ is a period of the exponential function).

Practically speaking, you can evaluate the expression $z^w$ by first writing $z = re^{it}$ and then using $$z^w = e^{w\log z} = e^{w(\ln r + it + 2k\pi i)}$$ for integral $k$. In your case, $z=-1$ and $w=-i$, so you can write $z = re^{it}$ with $r=1$ and $t=\pi$. Then you have the values $$(-1)^i = z^w = e^{w\log z}$$ $$ = e^{-i(\ln 1 + i\pi + 2k\pi i)}$$ $$ = e^{ -i(2k+1)\pi i}$$ $$= e^{(2k+1)\pi}$$ That is, the values are $$\ldots, e^{-5\pi}, e^{-3\pi}, e^{-\pi}, e^{\pi}, e^{3\pi}, e^{5\pi},\ldots$$ Taking $k=0$ give the principal value $e^{\pi}$.

MPW
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