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Okay this question can be interpreted as an extension to this question, however I need to have the validity of my claim checked.

Let $x \in \Bbb R, z \in \Bbb C, n \in \Bbb N$

The solution to the equation $$z^n = -x$$ is as follows:

Let $\omega \in \Bbb C$ such that $ \omega^n = -1$, Then

$$\omega^nz^n=(\omega z)^n=x.$$

We can define $\omega$ as the $n^{th}$ root solution to $-1$ as follows where $k, j \in \Bbb Z^+$ and $\,j, k<n$

$$\omega = \exp\left(\frac{2\pi k i }n\right)\exp\left(\frac{\pi i} n\right)$$

$$\implies \omega z = x^{1/n}$$then

$$z = \frac {x^{1/x}}{ \exp\left({2\pi k i }n^{-1}\right)\exp\left({\pi i} n^{-1}\right)}$$

We can define the $n^{th}$ roots of $x$ as

$$x^{1/n}=p_n\left(x\right)\exp\left(\frac{2\pi j i }n\right)$$

where $p_n(x)$ is the principle $n^{th}$ root of $x$ (as defined here). Finally

$$\begin{align} z&= p_n(x)\frac {\exp\left(2\pi j i n^{-1}\right)}{ \exp\left({2\pi k i }n^{-1}\right)\exp\left({\pi i} n^{-1}\right)}\\ z&=p_n(x)\,\,{\exp\left(i2\pi \left(j-k-2^{-1}\right)n^{-1}\right)} \end{align}$$

It can be stated that there are $n$ solutions using $j, k$ as selectors of these roots. If $j = k$ then it gives the same value for all values of $j$ and $k$ respectively.

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Ali Caglayan
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    By the functional equation of the exponential function, the quotient in the last equation reduces to $\exp (2\pi(j-k)in^{-1})$, and two pairs $(j_1,k_1), (j_2,k_2)$ give the same value if the remainder of the differences modulo $n$ is the same, so you have only $n$ distinct solutions (unless $z = 0$, when there is only one). – Daniel Fischer Aug 15 '13 at 21:20
  • What is the principle $n^{th}$ root of $x$? – Makoto Kato Aug 15 '13 at 22:16
  • @DanielFischer Thank you for the observation will change acordingly – Ali Caglayan Aug 15 '13 at 23:04
  • @MakotoKato the square roots of 1 are 1 and -1. The principle square root is 1. – Ali Caglayan Aug 15 '13 at 23:05
  • @Alizter What is the principle square root of $x = -2$? – Makoto Kato Aug 16 '13 at 21:18
  • @MakotoKato The square roots of $-2$ are $\sqrt 2 i$ and $- \sqrt 2 i$. The principle one is the first – Ali Caglayan Aug 16 '13 at 22:34
  • @Alizter Sorry. I should have asked what the principle 3rd root of -2 is. I hope you explain in general what the principle $n$-th root of $x$. – Makoto Kato Aug 16 '13 at 22:53
  • For $x \in \Bbb R^+$ the principle nth root is the nth root such that $z \in \Bbb R^+$. Also $x \in \Bbb R^- \implies z \in \Bbb R^-$ – Ali Caglayan Aug 16 '13 at 23:07
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    @Alizter Unfortunately, that description doesn't work. If $x\in\mathbb R^-$ and $n$ is even, then none of the $n$th roots of $x$ are in $\mathbb R^-$. Negative numbers are different from positive numbers, so you can't always take a statement about $\mathbb R^+$ and transplant it into $\mathbb R^-$. – Chris Culter Aug 16 '13 at 23:17

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Take a closer look at this part:

We can define $\omega$ as the $n^{th}$ root solution to $-1$ as follows where $k, j \in \Bbb Z^+$ and $\,j, k<n$

$$\omega = \exp\left(\frac{\pi k i }n\right)$$

What do you get if you calculate $\omega^n$?

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Chris Culter
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