# Power of a negative number

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:

$(-1)^{(\frac{22}{10})}$

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

$\sqrt[10]{(-1)^{22}}$

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 $(-1)^{(\frac{22}{10})}$?

• You might want to take a look at this: math.stackexchange.com/questions/317528/… – polynomial_donut Apr 17 '17 at 22:53
• 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/… – Mark Pineau Apr 17 '17 at 22:53
• The correct result can, of course, never depend on whether you cancel fractions. With your way of calculating, I can also "prove" that $-1=1$: $-1 = (-1)^1 = (-1)^{2/2} = \sqrt{(-1)^2}=\sqrt 1 = 1$ – 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.
HINT: $$-1=e^{i \pi}$$ Now, $$(-1)^\frac{22}{10}=e^{i \pi \frac{11}{5}}=\cos\frac{11 \pi}{5}+i \sin\frac{11 \pi}{5}=-\cos\frac{2 \pi}{5}-i \sin\frac{2 \pi}{5}$$
• Actually, $-1=e^{i\pi(2n+1)}, n\in\mathbb Z$. – celtschk Apr 17 '17 at 22:57