# Is $(-1)^{1/8} + (-1)^{7/8}$ ever a value whose real component is $0$?

Is

$$(-1)^{1/8} + (-1)^{7/8}$$

ever a value whose real component is $0$?

Is this ever true in modular arithmetic, hypercomplexes, and/or both?

• If a number has a "real component", that usually implies it also has an imaginary component. In modular arithmetic there's no such thing as a "real component". – Joe Z. Apr 16 '14 at 18:05

If I look in the complex plane at the eight numbers that when raised to the eighth power form $-1$, they are $\exp(\frac {2k+1}8\pi i)$ for $0 \le k \le 7, k$ integer. The seventh powers of these are the same set. I can certainly pick two that have a real part of their sum being $0$, for example $\exp(\frac {1}8\pi i)$ and $\exp(\frac {-1}8\pi i)=(\exp(\frac {9}8\pi i))^7$

Working modulo $2$, the eighth root of $-1$ is $1$. The seventh power of $1$ is $1$ and $1+1=0 \pmod 2$.

• LOL at the mod-2 solution. – Joe Z. Apr 16 '14 at 17:52
• @Joe Z: OP did ask about modular solutions. I tried some others first, but they didn't work. – Ross Millikan Apr 16 '14 at 18:21
• That's an interesting problem, actually. Are there any other systems besides $\mathbb{Z}_2$ where there exists that root? – Joe Z. Apr 16 '14 at 18:29

I'm pretty sure that in just the plain old complex numbers, $Re((-1)^{1/8} + (-1)^{7/8})$ is $\cos(\frac 18 \pi) + \cos(\frac 78 \pi) = 0$.

In fact, in general, if $a+b=1$, $(−1)^a+(−1)^b$ will have a real component of $0$.

However, you seem to be focusing on modular arithmetic and number systems, so I'll try to give you an answer in modular arithmetic as well.

In $\mathbb{Z}_n$ (the numbers modulo $n$), $(-1)^{1/8}$ doesn't really mean anything the way we usually talk about it, so we need to redefine it from first principles.

When we talk about $n^{1/k}$ or $\sqrt[k]n$, really what we mean is some number $m$ so that $m^k = n$.

So we define $(-1)^{1/8}$ as some number $m$ such that $m^8 \equiv -1 \pmod n$. Note that this number might not always exist — for example, in $\mathbb{Z}_3$, neither $1$ nor $2$ has an eighth power of $-1$. And there might be more than one — say, in $\mathbb{Z}_{17}$, the numbers $3, 5, 6, 7, 10, 11, 12, 14$ are all possible values of $(-1)^{1/8}$.

So we have $(-1)^{1/8}$, now we need $(-1)^{7/8}$. Well, going by the rules of exponents, we can just do $(-1)^{7/8} = ((-1)^{1/8})^7$, so supposing you still have that $m$ from above, $m^7$ will be your $(-1)^{7/8}$.

To recap, this means that in $\mathbb{Z}_n$, we're looking for a number $m$ that satisfies both $m^8 = -1$ and $m + m^7 = 0$, which fits your expression of $(-1)^{1/8} + (-1)^{7/8} = 0$.

As Ross Millikan pointed out in his answer, $\mathbb{Z}_2$ is one system in which this works — the number $1$ has $1^8 = 1$ and $1 + 1^7 = 0$.

But unfortunately, in $\mathbb{Z}_{17}$, none of those eight numbers we listed above satisfy both these equations. ($4$ and $13$ actually satisfy the latter one, but they're not eighth roots of $-1$.)

Try other values of $n$ and see in which ones a suitable $m$ exists.

If we are assuming that $(-1)^{7/8}=\left[(-1)^{1/8}\right]^7$, then no matter which eighth root of $-1$ we take, we get $$\mathrm{Re}\left[e^{i\pi(2k+1)/8}+e^{i\pi(14k+7)/8}\right]=0\tag{1}$$ The exponents sum to $i(2k+1)\pi$ which means the real part of the sum in $(1)$ is $0$ since $$\cos(x)+\cos(\pi-x)=0\tag{2}$$ Thus, no matter which eighth root of $-1$ we take, we get that $$\mathrm{Re}\left[(-1)^{1/8}+(-1)^{7/8}\right]=0\tag{3}$$