# Is there any notation for general $n$-th root $r$ such that $r^n=x$?

As we know that the notation for the $n$-th principal root is $\sqrt[n]{x}$ or $x^{1/n}$. But the principal root is not always the only possible root, e.g. for even $n$ and positive $x$ the principal root is always positive but there is also another negative root. E.g. consider $r^2=4$, then $\sqrt 4 =+2$, but $r=-2$ is also a valid solution. Since $x$ is a function of $r$ for some given $n$, so let

$$r^n=x=f(r).$$

We have $r=f^{-1} (x)$. Here $r \neq \sqrt[n]x$ because $\sqrt[n]x$ is the principal root not the general. So is there any notation like $\sqrt[n]{\phantom{aa}}$ for the general $n$-th root of the equation $r^n=x$?

• If $\zeta_n$ is a primitive $n$th root of the unity, then $\zeta^k_n\sqrt[n]{x}$ is an $n$th root of $x$ for any integer $0\leq k < n$. This are in fact all the $n$th roots of $x$ in the complex plane. You can define the multivalued function $f_k(x) := \zeta^k_n\sqrt[n]{x}$. Jun 30 '15 at 15:12
• Nope. Usually one says something like "let $r$ be an $n$-th root of $x$" or "let $r$ such that $r^n = x$". Also, note that you wrote $r = f^{-1}(x)$, but this doesn't make sense because in general $f$ isn't invertible.
– A.P.
Jun 30 '15 at 15:15
• @A.P. What is a non-invertible function? And why is $f$ not invertible here? Jun 30 '15 at 15:27
• A non-invertible function is one for which $f^{-1}$ is not defined. Usually "invertible" is the same as "bijective". In this case, $f$ may not be injective: for example $f(r) = r^2$ doesn't admit an inverse, i.e. an $f^{-1}$, because both $r$ and $-r$ have the same image.
– A.P.
Jun 30 '15 at 15:31

I'd say: $$r=z^{\frac{1}{n}}e^{\frac{2i\pi k}{n}}$$

It is a multivalued function with $k=0,\dots,n-1$

• Is $z$ a complex number here? And is $z^{1/n}$ multivalued in complex analysis or unique? Jun 30 '15 at 15:52
• $z$ is real in this case. If you have a complex number you can write is as $z=|z|e^{i\phi}$ and you'd have: $r=|z|^{\frac{1}{n}}e^{i\frac{\phi+2k\pi}{n}}$ Jun 30 '15 at 16:12

$$z^n=c\implies z=\omega^k\sqrt[n]c$$ Where $\omega$ is a primitive $n^{th}$ root of unity, and $0\le k\in\Bbb{Z}\le n-1$

For a non-negative real number $x$ there is always a unique choice of non-negative real $n$-th root, which is usually denoted by $\sqrt[n]{x}$. Furthermore, if $x$ is negative there is a unique choice of $n$-th root if $n$ is odd and none if $n$ is even.

In short, if $x \geq 0$ is real and $n$ is even, then the only real $n$-th roots of $x$ are $\pm \sqrt[n]{x}$ (and you can use this symbol), while if $x$ is real and $n$ is odd there is only one $n$-th root of $x$, denoted $\sqrt[n]{x}$.

You should understand, though, that taking roots usually involves a choice. In particular, every non-zero complex number has exactly $n$ $n$-th roots.

Now, you could keep the above choices for real numbers, but in general there is no canonical choice of root1. What we usually say instead is something like: "let $w$ be an $n$-th root of $z$". The nice thing, though, is that the other roots are then easily recoverable, because they are all of the form $$\zeta_n^i w \qquad \text{for } i \in \{0,\dotsc,n-1\}$$ where $\zeta_n$ is a primitive $n$-th root of unity, i.e. a complex number such that $\zeta_n^n = 1$ and $\zeta_n^m \neq 1$ for every $0 < m < n$. Again, a choice is involved here, but you can always take $$\zeta_n = e^{2\pi i/n}$$ TL;DR: If you wish to denote the generic $n$-th root of a complex number $z$ you may probably get away with the notation $z^{1/n}$. Just bear in mind that in general this is inherently ambiguous and you should treat this symbol more like a place-holder for an actual $n$-th root of $z$ than as a number.

[1] Technically, one could still define a unique choice of $n$-th root e.g. by taking the root with least argument (in $[0,2\pi)$). While this convention (or a similar one) may be used in analysis, I've never seen it in algebra or number theory.

In complex analysis, $\sqrt[n]{x}$ is regarded as a multivalued function. Or you can write it as $$\sqrt[n]{x}=\exp{\frac{\operatorname{Log}(x)}{n}},\space x\ne0.$$ $\operatorname{Log}(x)$ is the inverse function of $\exp(x)$, see here.

• Do you mean in complex analysis $\sqrt{4}=\pm 2$? Jun 30 '15 at 15:20
• Doesn't $\text{Log}$ take infinitely many different values? What I mean is: doesn't this just complicate matters?
– A.P.
Jun 30 '15 at 15:22
• @user103816 Yes, and no. $\sqrt[n]{x}$ is defined not on the complex plane, but on its Riemann surface. Jun 30 '15 at 15:25
• @A.P. Yes. But since the OP asked for a notation, only $\text{Log}$ is different from ordinary notations. Jun 30 '15 at 15:29
• @user103816 No, but after a fashion you could say that $4^{1/2} = \{2,-2\}$. The idea is to consider a covering space $p \colon S \to \mathbb{C}$ such that the pre-images of a complex number $z$ are precisely its square-roots (or, in general, its $n$-th roots).
– A.P.
Jun 30 '15 at 20:05

There is no privileged or standardized notation. It is a matter of definition, which is dependent on what is convenient for you. Sometimes you want to have a function that outputs an $n$-th root of the input, in which case you would define:

$a^b = \exp(b\ln_π(a))$ where $\ln_π$ is the principal branch of the natural logarithm.

This gives a well-defined function for any complex $a,b$ such that $\arg(a) \ne π$ (or equivalently $a$ is not on the non-positive real line). This function is differentiable everywhere in its domain, which is why it is often used. It is also continuous if a point on the negative real line is approached from the upper half-plane, and it coincides with the usual definition of exponentiation for positive real base and real exponent. Note that it does not coincide with the definition of exponentiation for negative real base and fractional real exponent. $(-8)^{\frac{1}{3}}$ is $(-2)$ in the 'real' world but is $(-2) \exp(\frac{2πi}{3})$ in the 'complex' world.

At other times we do not need a function that outputs a complex number but rather we work with sets of complex numbers. In that case we can define (for sets of complex numbers $a,b$:

$a^b = \exp(b\ln(a))$ where $ab = \{ zw : z \in a \land w \in b \}$ and $\exp(a) = \{ \exp(z) : z \in a \}$ and $\ln(a) = \{ z : \exp(z) \in a \}$.

If everything is suitably redefined to handle sets of complex numbers, we will get that this function is differentiable 'everywhere' except at $a = 0$. Not only that, some usual rules of real exponentiation hold. For example, $a^b a^c = a^{b+c}$ for any complex $a,b,c$ such that $a \ne 0$, and $a^c b^c = (ab)^c$ for any complex $a,b,c$ such that $ab \ne 0$. Some other rules still do not hold, such as $(\{3\}^{\{2\}})^{\{\frac{1}{2}\}} = \{3,-3\}$.