What is $(-1)^{\frac{1}{7}} + (-1)^{\frac{3}{7}} + (-1)^{\frac{5}{7}} + (-1)^{\frac{9}{7}} + (-1)^{\frac{11}{7}}+ (-1)^{\frac{13}{7}}$?

Question

The question is as given in the title. According to WolframAlpha, the answer is 1, but I am curious as to how one gets that. I tried simplifying the above into $$6(-1)^{\frac{1}{7}}$$ however that does not give 1.

Answer 1

The tagging seems to imply that this is basic exponents laws and the like, without any advanced stuff, so by going by the definitions: if $\;n,\,m\;$ are odd and coprime, then

$$(-1)^{m/n}=\left((-1)^{1/n}\right)^m=(-1)^m=-1$$

and also

$$(-1)^{m/n}=\left((-1)^m\right)^{1/n}=(-1)^{1/n}=-1$$

and thus your expression is just simply $\;(-1)+(-1)+\ldots+(-1)=-6 \;$ .

Answer 2

Notice: $$-1=\left|-1\right|e^{\left(\arg\left(-1\right)+2\pi\text{k}\right)i}=e^{\pi\left(1+2\text{k}\right)i}$$ Where $\text{k}\in\mathbb{Z}$

So, when $\text{n}\in\mathbb{R}$:

$$\left(-1\right)^\text{n}=e^{\text{n}\pi\left(1+2\text{k}\right)i}=\cos\left(\text{n}\pi\left(1+2\text{k}\right)\right)+\sin\left(\text{n}\pi\left(1+2\text{k}\right)\right)i=\cos\left(\text{n}\pi\right)+\sin\left(\text{n}\pi\right)i$$

So (they are called the principal roots):

1. When $\text{n}=\frac{1}{7}$: $$\left(-1\right)^{\frac{1}{7}}=\cos\left(\frac{\pi}{7}\right)+\sin\left(\frac{\pi}{7}\right)i$$
2. When $\text{n}=\frac{3}{7}$: $$\left(-1\right)^{\frac{3}{7}}=\cos\left(\frac{3\pi}{7}\right)+\sin\left(\frac{3\pi}{7}\right)i$$
3. When $\text{n}=\frac{5}{7}$: $$\left(-1\right)^{\frac{5}{7}}=\cos\left(\frac{5\pi}{7}\right)+\sin\left(\frac{5\pi}{7}\right)i$$
4. When $\text{n}=\frac{9}{7}$: $$\left(-1\right)^{\frac{9}{7}}=\cos\left(\frac{9\pi}{7}\right)+\sin\left(\frac{9\pi}{7}\right)i$$
5. When $\text{n}=\frac{11}{7}$: $$\left(-1\right)^{\frac{11}{7}}=\cos\left(\frac{11\pi}{7}\right)+\sin\left(\frac{11\pi}{7}\right)i$$
6. When $\text{n}=\frac{13}{7}$: $$\left(-1\right)^{\frac{13}{7}}=\cos\left(\frac{13\pi}{7}\right)+\sin\left(\frac{13\pi}{7}\right)i$$

Adding those together, gives us:

$$2\left\{\cos\left(\frac{\pi}{7}\right)+\sin\left(\frac{\pi}{14}\right)-\sin\left(\frac{3\pi}{14}\right)\right\}=1$$

Answer 3

Note that $(-1)^{\frac{1}{7}} = -1$

Also note that $(-1)^{\frac{9}{7}} = -(-1)^{\frac{2}{7}},\ (-1)^{\frac{11}{7}} = -(-1)^{\frac{4}{7}},\ (-1)^{\frac{13}{7}} = -(-1)^{\frac{6}{7}}$

You can then rewrite your addition as

$$(-1)^{\frac{1}{7}} - (-1)^{\frac{2}{7}} + (-1)^{\frac{3}{7}} - (-1)^{\frac{4}{7}} + (-1)^{\frac{5}{7}} - (-1)^{\frac{6}{7}}$$

Now remembering that $a^{\frac{p}{q}} = (a^{\frac{1}{q}})^p = (\sqrt[q]{a})^p$ and substituting $q = 7$ and the appropriate $p$ you get:

$$(-1)^{\frac{1}{7}} - (-1)^{\frac{2}{7}} + (-1)^{\frac{3}{7}} - (-1)^{\frac{4}{7}} + (-1)^{\frac{5}{7}} - (-1)^{\frac{6}{7}} \\ = (-1) - (1) + (-1) - (1) + (-1) - (1) \\ = -6$$

Now what if instead of $a^{\frac{p}{q}} = (a^{\frac{1}{q}})^p$ you make $a^{\frac{p}{q}} = \sqrt[q]{a^p}$? Then for negative $a$, odd $p$ and even $q$ that would only be calculatable if you entered the realm of the complex numbers and made use of the imaginary unit. But when you enter the realm of the complex numbers, you get that $\sqrt[q]{x}$ has $q$ different values. That is, there are $q$ different numbers $x_i$ for which $(x_i)^q = x$.

WolframAlpha goes straight ahead and uses the complex plane to calculate the roots making use of the complex numbers. Now if you make use of the fact that the sum of all $q$ roots of $x$ is 0, you can do the following:

Call $x_i$ the ith root of (-1). Then $x_i = (-1)^{\frac{i}{7}}$ and if you notice that in your sum all roots are represented except for $x_0$ which is -1, then

$$\sum_{i=0}^{6} x_i = -1 + \sum_{i=1}^{6} x_i$$ but $$\sum_{i=0}^{6} x_i = 0$$ so $$0 = -1 + \sum_{i=1}^{6} x_i \iff \sum_{i=1}^{6} x_i = 1$$

Answer 4

We have $$x^7+1=(x+1)(x^6-x^5+x^4-x^3+x^2-x+1).$$ Let $x=(-1)^{1/7}$ and get the result.