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It is well known that $x^2-1=0$ has two roots in $\mathbb{C}$, namely $\pm 1$. In general $x^n-1=0$ has exactly $n$ roots in $\mathbb{C}$. But what happens when $n$ is non integer (rational or real or even complex)? For example how many roots does $x^{1.9}-1=0$ have (counting multiplicity)? Intuitively I suspect that for adequately small $\epsilon$, $x^{1+\epsilon}-1=0$ has one complex root while $x^{2-\epsilon}-1=0$ (for some other $\epsilon$) will have 2 roots and more I guess that these roots will be close to $\pm 1$ respectively. Note that multiplicity of a root $\rho$ of a (non-polynomial) function $f$ is defined as the largest integer $k$ (if exists) such that:

\begin{align*} \lim_{x \rightarrow \rho} \frac{|f(x)|}{|x-\rho|^k} < \infty \end{align*}

Update 1: I understand that my claim regarding the number of roots in $\mathbb{C}$ is not correct since $x^2=1$ has two complex roots but $x^{2.1}=1$ has a lot. My question now is whether the following holds:

Claim 1: There is an $\delta \in \mathbb{R}$ such that for all $\epsilon \in \Re$ with $|\epsilon|<\delta$ it holds that $x^{2+\epsilon}=1$ has two real roots (as many as the initial one).

Note that this claim is only about the number of real roots of the equation. Some simulations on MATLAB show that this might be true. Additionally, the simulations show that the real roots of the perturbed equation are close to the roots of the unperturbed. This claim, if it holds, can easily be extended to cater for equations with more exponents.

Update 2: Consider for example the equation $x^{1.5}=1$. This has solutions $\rho_i=\exp\left(4k\pi i/3\right)=\cos(4k\pi/3)+i \sin(4k\pi/3)$ with $k\in\mathbb{Z}$. What is the multiplicity of each root??? According to the definition, it should be $1$, so I think it would be good when saying that the solutions of $x^a=1$ are $\exp(2k\pi i /a)$ to restrict $k$ properly so that $2k\pi / a \in [0,2\pi)$.

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    Note that $x^{0.5}$ is the same as the square root function, which is has two values.2011-03-22

2 Answers 2

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As pointed out in the comments, taking the logarithm of both sides of $ x^a = 1, $ and taking into account the multi-valued nature of the logarithm, gives $ a \log x = 2 \pi k i $ for some integer $k$, or $\log x = 2\pi k i/a$ (assuming $a\neq 0$). Then exponentiating both sides of this equation gives $ x = \exp\left(\frac{2\pi k i}{a}\right) = \cos\left(\frac{2\pi k}{a}\right) + i\sin\left(\frac{2\pi k}{a}\right). $ If $a$ is rational and equal to $p/q$ in lowest terms, then this takes on exactly $p$ different values; otherwise it takes on infinitely many different values, dense on the unit circle.

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    @mjqxxxx A minor observation: for $\alpha=N/\pi$ with $N\in\mathbb{N}$, $s^{\alpha}=1$ has finitely many distinct roots on the unit circle.2014-12-08
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The ONLY entire functions that has finitely many zeros are functions that are/(can be written in) the form $P(z) e^{g(z)}$ where $g$ is entire.

This might answer your question partly, I suspect.