I know this may be silly but lately I've been wondering about powers of negative numbers and I came up with an ambiguous example:


I wanted to know if it was $+1$ or $-1$. My first attempt was writing it down as a root:


Minus one to the even power is obviously plus one, then 1 to the power of $\frac{1}{10}$ is 1. Yet if we first reduce the fraction and then calculate we get a different answer:

$(-1)^{(\frac{22}{10})} = (-1)^{(\frac{11}{5})} = \sqrt[5]{(-1)^{11}}$

Even though it's (I suppose) the same situation, $(-1)^{11}$ is $-1$ and we end up with $\sqrt[5]{-1}$, a complex number. What is (or is there even) the value of $(-1)^{(\frac{22}{10})}$?

  • $\begingroup$ You might want to take a look at this: math.stackexchange.com/questions/317528/… $\endgroup$ – polynomial_donut Apr 17 '17 at 22:53
  • $\begingroup$ In addition to Uddeshya Singh's answer, if you want a detailed explanation as to where the equation $e^{i\pi}=-1$ comes from, check this link out. math.stackexchange.com/questions/2223267/… $\endgroup$ – Mark Pineau Apr 17 '17 at 22:53
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    $\begingroup$ The correct result can, of course, never depend on whether you cancel fractions. With your way of calculating, I can also "prove" that $-1=1$: $-1 = (-1)^1 = (-1)^{2/2} = \sqrt{(-1)^2}=\sqrt 1 = 1$ $\endgroup$ – celtschk Apr 17 '17 at 22:53

$(a)^{bc}= a^ba^c$ is not an equation that holds for all $a,b,c$. Although that equation is true when all of $a,b,c$ are positive real numbers, as you've discovered it's not true in general.

The correct approach is generally considered to simplify the exponent and obtain the complex number, though there are actually three different ways you can think of exponents with a negative base, as discussed here.


HINT: $$-1=e^{i \pi}$$ Now, $$(-1)^\frac{22}{10}=e^{i \pi \frac{11}{5}}=\cos\frac{11 \pi}{5}+i \sin\frac{11 \pi}{5}=-\cos\frac{2 \pi}{5}-i \sin\frac{2 \pi}{5}$$

Hope you got your answer?

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    $\begingroup$ Actually, $-1=e^{i\pi(2n+1)}, n\in\mathbb Z$. $\endgroup$ – celtschk Apr 17 '17 at 22:57
  • $\begingroup$ I got it, thank you very much! $\endgroup$ – Virginia Apr 17 '17 at 22:58

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