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 :)

  • 1
    $\begingroup$ $$(-1)^a=e^{a\log(-1)}=e^{a(i\pi +i2n\pi)}=e^{ia\pi (2n+1)}=\cos(a\pi (2n+1))+i\sin(a\pi (2n+1))$$ $\endgroup$
    – Mark Viola
    Aug 12, 2016 at 3:34
  • $\begingroup$ 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. $\endgroup$
    – fleablood
    Aug 12, 2016 at 4:54
  • $\begingroup$ Note that $(-1)^{\frac{a}{b}} \neq \frac{(-1)^a}{(-1)^b}$! $\endgroup$
    – KoA
    Aug 12, 2016 at 5:00
  • $\begingroup$ Duplicate of: math.stackexchange.com/questions/1749096/1-sqrt2 $\endgroup$
    – N.S.JOHN
    Aug 12, 2016 at 17:48
  • $\begingroup$ We never seem to run out of these complex exponent type of questions, now don't we? $\endgroup$ Aug 17, 2016 at 0:12

1 Answer 1


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 $\text{Arg}(z)=\pi$.

  • $\begingroup$ Does n in this case represent a "counter" that cycles around $\sin{x}$ and $\cos{x}$? $\endgroup$
    – Ian L
    Aug 12, 2016 at 3:40
  • $\begingroup$ $n$ is any integer that fixes the Riemann Sheet of the complex logarithm. $\endgroup$
    – Mark Viola
    Aug 12, 2016 at 3:41
  • $\begingroup$ What math do I need to learn in order to learn about the complex logarithm? Also, enjoy the Olympics :) $\endgroup$
    – Ian L
    Aug 12, 2016 at 3:42
  • $\begingroup$ @IanLimarta A introductory course in Complex Variables will include the complex logarithm. - Mark $\endgroup$
    – Mark Viola
    Aug 12, 2016 at 3:44
  • $\begingroup$ As I thought: some type of complex analysis. Like usual ): $\endgroup$
    – Ian L
    Aug 12, 2016 at 3:46

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