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 $2k\pi / a \in [0,2\pi)$.

  • 4
    $\begingroup$ if $a$ is not an integer then $x^a$ is a multi-valued function ($x^a=\exp(a\log x)$ for all complex values of $\log x$). The solutions of your equations are $\exp(2k\pi i/a)$ for all integers $k$, and that's an infinite set if $a$ is irrational. $\endgroup$
    – user8268
    Commented Mar 22, 2011 at 16:09
  • $\begingroup$ @user8268: Thanks for the answer! It's very interesting but could you give me a reference so that I can read more about these equations? Is there any chance to solve analytically $x^2.1+x^0.9-3=0$ ? $\endgroup$ Commented Mar 22, 2011 at 16:21
  • $\begingroup$ Note that $x^{0.5}$ is the same as the square root function, which is has two values. $\endgroup$ Commented Mar 22, 2011 at 16:29

2 Answers 2


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.

  • $\begingroup$ Ok, a few points here: First, where do I find a book to read more? Second, what happens with more complex equations (like $x^2.1+x^0.9+1=0$ ). How do I solve them? Third, does it hold that for every choice of $a_0,a_1,\ldots,a_n \in \Re$ there is $\epsilon>0$ so that for all $0<\epsilon_1, \epsilon_2, \ldots, \epsilon_n < \epsilon$ the equation $a_0+a_1x^{1+\epsilon_1}+...+a_nx^{n+\epsilon_n}=0$ has as many REAL roots as $a_0+a_1x+...+a_nx^n=0$ ? I see for example that this holds for $x^2=1$ and $x^2.05=1$ $\endgroup$ Commented Mar 22, 2011 at 18:25
  • $\begingroup$ 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. A reference for this claim??? $\endgroup$ Commented Mar 22, 2011 at 22:38
  • $\begingroup$ @Pantelis: I suspect that shifting the exponents slightly can change the number of real roots by splitting degenerate roots. For instance, $x^2-2x+1=0$ has one real root ($x=1$), but presumably $x^{2+\epsilon_2} - 2x^{1+\epsilon_1} + 1 = 0$ has two real roots for small positive $\epsilon_1$ and $\epsilon_2$ where $\epsilon_2 \neq 2\epsilon_1$. $\endgroup$
    – mjqxxxx
    Commented Mar 24, 2011 at 18:26
  • $\begingroup$ Ok, I agree. But the equation you chose had from the beginning (unperturbed version) 2 real root that happened to be the same number (multiplicity=2). So, the claim regarding the number of roots might be true... $\endgroup$ Commented Mar 25, 2011 at 7:25
  • 1
    $\begingroup$ @Pantelis: The fact that $e^{\frac{2\pi k}{a}i}$ with irrational $a$ is dense on the unit circle is a consequence of irrationality of $\pi$ (we were talking about this here in the comments). You can find an easy explanation of this in Needham's Visual complex analysis, 3.VI: "Multifunctions" (especially the subsection "General powers"). Hope this helps $\endgroup$ Commented Mar 26, 2011 at 0:59

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .