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We know that the following conditions are true:

$(-1)^{2n}$ is $1$ where $n$ is an integer

$(-1)^{2n+1}$ is $-1$ where $n$ is an integer.

We can extend this reasoning for rational numbers.

If we let a number be written in the form of $a+\frac{b}{c}$ such that $\gcd(b,c) = 1$ and $a, b, c$ are integers, then to evaluate $(-1)^{a+\frac{b}{c}}$, we can separate $(-1)^{a+\frac{b}{c}}$ into $(-1)^{a}\times\frac{(-1)^b}{(-1)^c}$. The terms $(-1)^{a}$, $(-1)^b$, $(-1)^c$ can be reduced to either of the two cases above; and we solve for $-1$ or $1$.

Now what is $(-1)^\pi$. It isn't rational, so there isn't a perfect ratio that can be done with the method above. I'm suspecting that roots of unity come to play here or DeMoirve's Theorem, but don't they resolve to whether or not you can solve $r^n$ where r is the radius of the complex number and n is exponent.

NOTE: I'm still in high school, so I am looking for an answer that does not have any advance math- but will gladly accept any response to the question for curiosity's sake.

Thank you :)

JMoravitz
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Ian L
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    $$(-1)^a=e^{a\log(-1)}=e^{a(i\pi +i2n\pi)}=e^{ia\pi (2n+1)}=\cos(a\pi (2n+1))+i\sin(a\pi (2n+1))$$ – Mark Viola Aug 12 '16 at 03:34
  • If b > 0. b^x is continuous on Q, the rationals. (Continuous is an advanced concept but it means if x and y are close b^x and b^y are close.) So we can extend b^x, x irrational by taking limits. (Also an advanced topic but easy to understand intuitively). If b <0 then b^r, r rational is not continuous and jumps between positive, negative and undefined like a flea. So b^x makes no sense. Until we study complex numbers. Which no matter who you look at it is very advanced. Then the definition becomes something looking very different. – fleablood Aug 12 '16 at 04:54
  • Note that $(-1)^{\frac{a}{b}} \neq \frac{(-1)^a}{(-1)^b}$! – KoA Aug 12 '16 at 05:00
  • Duplicate of: http://math.stackexchange.com/questions/1749096/1-sqrt2 – N.S.JOHN Aug 12 '16 at 17:48
  • We never seem to run out of these complex exponent type of questions, now don't we? – Simply Beautiful Art Aug 17 '16 at 00:12

1 Answers1

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For any complex numbers $z$ and $a$, the term $z^a$ is defined as $z^a=e^{a\log(z)}$. Then, with $z=-1$ and $a=\pi$ we have

$$\begin{align} (-1)^\pi&=e^{\pi\log(-1)} \tag 1\\\\ &=e^{\pi(i\pi +i2n\pi)} \tag 2\\\\ &=e^{i\pi^2 (2n+1)}\\\\ &=\cos(\pi^2 (2n+1))+i\sin(\pi^2 (2n+1)) \end{align}$$

for all integer values of $n$.

Note that in going from $(1)$ to $(2)$ we recognized the multivalued nature of the complex logarithm, $\log(z)=\log(|z|)+i(\text{Arg}(z)+2n\pi)$. Here, $|z|=1$ and $\text{Arg}(z)=\pi$.

Mark Viola
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  • Does n in this case represent a "counter" that cycles around $\sin{x}$ and $\cos{x}$? – Ian L Aug 12 '16 at 03:40
  • $n$ is any integer that fixes the Riemann Sheet of the complex logarithm. – Mark Viola Aug 12 '16 at 03:41
  • What math do I need to learn in order to learn about the complex logarithm? Also, enjoy the Olympics :) – Ian L Aug 12 '16 at 03:42
  • @IanLimarta A introductory course in Complex Variables will include the complex logarithm. - Mark – Mark Viola Aug 12 '16 at 03:44
  • As I thought: some type of complex analysis. Like usual ): – Ian L Aug 12 '16 at 03:46
  • @IanLimarta Yes. Note that the answer is indeed complex. – Mark Viola Aug 12 '16 at 03:48
  • @Giffyguy Actually, WA gives THIS ANSWER, which corresponds to $n=0$. -Mark – Mark Viola Aug 12 '16 at 03:53
  • @IanLimarta Its actually explainable once you learn Taylor's Theorem, which you probably don't understand, but if you look at some of its example formulas, they can be quite interesting. (On some Wikipedia would do) – Simply Beautiful Art Aug 17 '16 at 00:14
  • @SimpleArt What do you mean? This topic is unrelated to Taylor's expansion. – Mark Viola Aug 17 '16 at 01:17
  • Even if it does not apply, isn't Taylor's Theorem the approximation equation for a differentiable function compared to a Taylor polynomial? – Ian L Aug 17 '16 at 18:45
  • @IanLimarta But if $a^x$ is analytic (you could google that term if you don't know what it means), then Taylor's theorem can go somewhere further than mere approximation. And the same with the cosine and sine, which quite magically become equal around $a=0$. With that much, I was thinking you don't really need to learn complex analysis, to understand the problem/solution to a comfortable degree. Requires some calculus though. – Simply Beautiful Art Aug 17 '16 at 21:12
  • @simpleart This exponential involves a negative number raised to a non-integer number. How is that defined without invoking complex numbers? And what on earth does Taylor have to do with its evaluation? – Mark Viola Aug 17 '16 at 21:17