According to the standard definition of a real-valued root, equations $\sqrt{x}=2$, $\sqrt[3]{x}=2$ and $\sqrt[3]{x}=-2$ have real-valued solutions, but $\sqrt{x}=-2$ doesn't. But when we move to the complex world, complex-valued roots become essentially multivalued, so we (potentially) should be able to solve more equations over $\mathbb{C}$. So do (or doesn't) equations $\sqrt{x}=-2$ and $\sqrt[3]{x}=-2$ have complex solutions and which definitions of the complex-valued root should we use to obtain them.


3 Answers 3


I think it is a good idea to solve your first equation $\sqrt{x}=-2$ from first principles:

We know that any complex number different from zero has a polar representation:

$$x = re^{i\theta}$$

Therefore, by the De Moivre's theorem:

$$\sqrt{x} = \sqrt{r}\left(\cos\left(\frac{\theta+2\pi k }{2}\right) + i \sin\left(\frac{\theta+2\pi k }{2}\right) \right) \quad k=0,1$$

Suppose $k=0$.

Since $|\sqrt{x}|=\sqrt{r}$ and $|-2|=2$ we should have $\sqrt{r} = 2$.

So, our equation $\sqrt{x}=-2$ is equivalent to solve:

$$2\left(\cos\left(\frac{\theta}{2}\right) + i \sin\left(\frac{\theta}{2}\right) \right) = -2$$

or what is the same:

$$\cos\left(\frac{\theta }{2}\right) + i \sin\left(\frac{\theta }{2}\right) = -1$$

From the previous relation we get this system of equations:

$$ \cos\left(\frac{\theta }{2}\right) = -1 \tag{1}$$ $$ \sin\left(\frac{\theta }{2}\right) = 0 \tag{2} $$

Thus, from (1):

$$ \cos\left(\frac{\theta }{2}\right) = -1 \implies \theta = 2\pi +4\pi n \quad n\in \mathbb{Z} $$

Substituting in (2):

$$ \sin\left(\frac{2\pi +4\pi n}{2}\right) = \sin(\pi+2\pi n)= 0 $$

and we have obtained our first solution

$$ x^* = 4e^{i(2\pi+4\pi n)} =4 $$

Now suppose that $k=1$

We need to solve the system:

$$ \cos\left(\frac{\theta}{2}+\pi \right) = -1 \tag{3}$$ $$ \sin\left(\frac{\theta }{2}+\pi\right) = 0 \tag{4} $$

From (3):

$$\theta = \pi + 2\pi n \quad n\in \mathbb{Z} $$

Substituting in (4):

$$ \sin\left(\frac{\pi + 2\pi n }{2}+\pi\right) = \sin\left(\frac{\pi}{2}+\pi(n+1)\right) = (-1)^{n+1} $$

which is never zero and clearly contradicts our assumption in (4).

So, there are no solutions if $k=1$.

So what it is all about?

It turned out that if $z=re^{i\theta},\; r>0 $ the function $\sqrt{z}$ is multivalued:

$$ \sqrt{z} = \begin{cases} \displaystyle \sqrt{r}\left(\cos\left(\frac{\theta }{2}\right) + i \sin\left(\frac{\theta }{2}\right) \right) \\ {} \\\displaystyle \sqrt{r}\left(\cos\left(\frac{\theta }{2}+\pi \right) + i \sin\left(\frac{\theta }{2}+\pi \right) \right) \end{cases}$$

and depending on the $k=0,1$ you choose the equation has a solution or not.

Why the software only choose one of the values for k?

  1. The first reason is simply for comfort, to make the function single valued.
  2. The second reason is that if in addition to have a single-valued function we restrict the domain to, for example, $-\pi <\theta<\pi$ we also have an analytic and continuous function.

IMO it is ok to try to solve equations using multi-valued functions. You just need to specify which root $k=0,...n-1$ you are using to define your function $\sqrt[n]{z}$. However, consider that to solve $\sqrt[n]{x} = b$ the complexity of your task increases considerably as $n$ increases.

Edit:Is the equation $\mathbf{\sqrt[n]{z} = b \quad n\in \mathbb{N}}$ always solvable on $\mathbf{\mathbb{C}}$.

The answer is yes.

Note that if we put

$$\displaystyle \varphi = \frac{\theta}{2},\quad \phi = \frac{\theta}{2}+\pi, \quad |\theta|<\pi $$

the image of each branch of the multivalued function $\sqrt{z}$ almost divides the complex plane in two:

$$ f_0(z) = \sqrt{r}\left(\cos\varphi + i \sin\varphi \right) \quad -\frac{\pi}{2} <\varphi <\frac{\pi}{2} $$

$$ f_1(z) = \sqrt{r}\left(\cos\phi + i \sin\phi \right) \quad \frac{\pi}{2} <\phi < \frac{3\pi}{2} $$

if we allow $f_0(z)$ to attain the value $\theta = \pi$, or what is the same $\displaystyle \varphi = \frac{\pi}{2}$ and we define $f_0(0) = 0$. Then $f_0(z)$ and $f_1(z)$ cover the whole complex plane. However, we lost the continuity of one branch, $f_0(z)$.

Similarly, the image of each branch of the multivalued function $\sqrt[n]{z}$ can be defined to divide the complex plane in $n$ regions, the union of all being the whole complex plane. This also at the expense of the continuity.

So, to solve $\sqrt[n]{z} = b$ we only need to know which of the $n$ regions $b$ belongs to.

  • $\begingroup$ So with this approach we essentially generalize our equations $\sqrt[n]{z}=a$ to inclusions $\sqrt[n]{z}\ni a$ or rather $\sqrt[n]{z}-a \ni 0$. $\endgroup$ Commented May 17 at 22:45
  • 1
    $\begingroup$ @Buckminster Hi, I 'm not sure what the correct notation should be. Actually, the name multivalued-functions is misleading since the "multivalued-function" we defined as the union of $n$ branches is not a function at all, it is a surjective binary relation, that is the technical name. $\endgroup$
    – Bertrand87
    Commented May 18 at 1:22

$\sqrt{x} = -2$ doesn't have a solution in $\mathbb{R}$ because when we define the real-valued square-root function $\sqrt{\cdot}$ to be the inverse of the square function $x\mapsto x^2$, we choose to define it as the positive real number whose square is equal to the input value. So by definition there's no solution in $\mathbb{R}$ for $\sqrt{x} = -2$.

So it is not a matter of field extension we are talking about. The complexe numbers were introduced so that the negative numbers have square roots, not that the negative numbers have squares. Negative numbers always have squares. They are just not chosen by convention to be the output value of the square-root function.

If $\sqrt{x} = -2$ has a solution in $\mathbb{C}$, by squaring both sides we have $x = 4$. And this means we redefine the square-root function $\sqrt{\cdot}$ in $\mathbb{C}$, which is not the case in practice and not good. The same goes for $\sqrt[3]{x} = -2$.

  • 1
    $\begingroup$ "which is not the case in practice and not good" -- sound like an ancient Roman law: ars mathematica damnabilis et interdicta est. BTW, I'm not convinced by your answer since $x^2-4=0$ has two solutions (both not complex). So $\sqrt{4}=2$ by definition while when solving $x^2-4=0$ it looks different. Thus $\sqrt{4} = \pm 2$ at times. $\endgroup$
    – m-stgt
    Commented May 15 at 14:20
  • $\begingroup$ @m-stgt You certainly can define the square-root function the way you like, but as I said, it is rare in practice and "not good". And once you define it that way, it remains true in $\mathbb{C}$ too. So $\sqrt{x} = -2$ will have solution as $x = 4$. And that is the only one because if you have $\sqrt{x} = -2$, by squaring both sides you always get $x = 4$. $\endgroup$
    – corindo
    Commented May 15 at 14:24
  • $\begingroup$ @corindo > The same goes for $\sqrt[3]{x}=-2$ But is it really the same? In the real world $\sqrt[3]{-8}=-2$, so we are using one and the same cube root notation for different things already. Can't we then just say that $\sqrt[3]{-8}=\{-2;1\pm i\sqrt{3}\}$ and call it a day, so that every irrational equation can have a root over $\mathbb{C}$? $\endgroup$ Commented May 16 at 0:16
  • $\begingroup$ @corindo > this means we redefine the square-root function $\sqrt{\cdot}$ in $\mathbb{C}$, which is not the case in practice But is it really not the case in practice? I suspect there should be some generalizations of an equation suitable for multivalued functions. That was the main point of my question: are there such generalizations of an equation that may allow us to always find solutions to an arbitrary irrational equation over $\mathbb{C}$, where a complex nth root is essentially multivalued. $\endgroup$ Commented May 16 at 7:15

$\newcommand{\Sqrt}{\sqrt{\rule{0pt}{4pt}\quad}}$tl; dr 1: Reasonable as it sounds, the question as currently worded ("Do equations $\sqrt{x} = -2$ and $\sqrt[3]{x} = -2$ have complex solutions?" without further context) is not well-posed.

tl; dr 2.: When people think of "being able to solve more equations over the complex numbers," it usually refers not to having square roots that are negative, but to being able to take square roots of negative reals (and more).

There are at least two standard ways to interpret the face-value question over the complex numbers:

  1. Square both sides (or for the second, cube both sides) and interpret the question as asking for solutions of $x = (-2)^{2} = 4$ (or of $x = (-2)^{3} = -8$).
  2. Fix a function $\Sqrt$, including its domain (and codomain), and ask about that specific function.

The first is ... probably disappointing: The process of interpreting hands us the unique solutions, and doesn't involve complex numbers (or root functions) at all.

As for the second, the answer depends on the function we pick, but in the end is perhaps disappointing. For example, there are two "standard" choices of complex square root function:

  • The complex-valued principal branch, which on the non-negative reals agrees with the real (non-negative) square root, and on the negative reals is either undefined, or discontinuous with positive imaginary part;
  • The non-principal negative of the principal branch. Over the non-negative reals this is denoted $-\Sqrt$. (There are other choices of square root function, but none differs essentially from these for purposes of this question.)

As in corindo's (+1) answer, the equation $\sqrt{x} = -2$ has no solution for the principal branch, and has solution $x = 4$ for the non-principal branch. The end result is, we've brought in heavy machinery and found ... the disappointing answer we got by squaring in item 1. (The same thing happens for complex cubing, so long as we define the principal cube root to be negative real on the negative real axis. Details are left as an exercise.)

Added: In the comments I claimed, "Loosely, multi-valuedness of $f$ doesn't give more solutions $z$ of $f(z) = c$, but fewer. (!!)" To explain the intent, let's place this question in a moderately general context: Assume $F$ is a non-constant entire function (holomorphic on the complex plane), and let $f$ be a (single-valued) branch of inverse, defined in some non-empty (probably open) set $U$.

For example, perhaps $F(w) = w^{2}$ is the squaring map, and $f$ is a branch of square root:

The images of the principal square root and non-principal square root The squaring map, whose domain may be viewed as the Riemann surface of the multi-valued square root

To say we have a branch of inverse means $F\bigl(f(z)\bigr) = z$ for all $z$ in $U$. Applying the single-valued function $f$ to both sides and writing $f(z) = c$ gives $f\bigl(F(c)\bigr) = c$ for every value $c$ of $f$.

The gist of my comment is, "many" complex $c$ are not values of $f$: While an entire function such as $F$ is "almost surjective" (the image omits at most one complex value by the "big Picard theorem"), a branch of inverse such as $f$ is "far from being surjective" (unless $F$ is actually invertible). In the diagrams, the image of the principal square root is the right half-plane (including the positive imaginary axis if we define the square root on the negative reals), and the image of the non-principal square root is the left half-plane (with the negative imaginary axis).

In other words, suppose we have a holomorphic function $f$ defined on some non-empty open set, $c$ is fixed, and we want to count solutions $z$ to $f(z) = c$. If we know $f$ is entire, we are assured there is at least one solution (unless $f$ happens to omit $c$ as allowed by big Picard), and unless $f$ is a linear polynomial there are multiple solutions, perhaps infinitely many.

If, by contrast, $f$ is a branch of inverse of an entire function, we expect $f(z) = c$ to have no solutions. In detail, if we consider a "complete collection of branches of inverse" (a non-standard term meaning distinct branches have disjoint images and the union of the images is the whole complex plane), then among the complete collection of branches there is precisely one solution $z$, namely $z = F(c)$, with $f(z) = c$ for precisely one branch!


It's true that (for example) $\sqrt{z} = -2$ has no complex solutions if the square root is the principal branch, while there is a solution if we consider the non-principal branch. The main points of this answer are:

  • If $F$ is entire and surjective (e.g., an arbitrary polynomial, by the fundamental theorem of algebra), $c$ is complex, and if $f$ is a branch of inverse of $F$, then the equation $f(z) = c$ has at most one solution, $z = F(c)$. Further,
  • Among all branches of inverse in a "complete collection," only one branch admits a solution $z$ of $f(z) = c$.
  • $\begingroup$ The question was "which definitions of the complex-valued nth root should we use to obtain solutions (to arbitrary irrational equations over the field of complex numbers where a complex root is essentially multivalued)". We somehow need to check/verify the solutions we obtained, and for that purpose we need to choose an appropriate definition of a complex root (and probably also a definition/generalization of an equation suitable for multivalued functions). $\endgroup$ Commented May 16 at 7:11
  • $\begingroup$ The domain of a single-valued $n$th root function $f$ is crucial, not just the formula for a complex $n$th root (or more general relation): The domain determines the image (set of values), which by definition characterizes for which complex $c$ the equation $f(z) = c$ has a solution $z$. <> Separately, if $c \neq 0$, then with an arbitrary choice of domain, for at most one branch $f$ of $n$th root is there a solution $z$ of $f(z) = c$, and that solution is $z = c^{n}$, as under item 1. <> Loosely, multi-valuedness of $f$ doesn't give more solutions $z$ of $f(z) = c$, but fewer. (!!) $\endgroup$ Commented May 16 at 13:53
  • 1
    $\begingroup$ Can you please elaborate a bit more on you last statement? With multi-valuedness of a complex $n$th root we can solve both $\sqrt{z}=-2$ (which we couldn't with either the definition of a real-valued root or a principal complex square root) and $\sqrt[3]{z}=-2$ (which we also couldn't solve with the most "popular" definition of a principal complex square root) . So how can there be fewer solutions? $\endgroup$ Commented May 16 at 16:34

You must log in to answer this question.

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