The set of nth roots of a complex number z contains all complex numbers w such that wn = z. If we use the polar notation for complex numbers, so that z = Rei θ and w = Sei φ, then we find there are n roots of the form:

S = R1/n and φ = θ/n + 2π(k/n) where k = 0, 1, ..., n-1


When dealing with the real numbers, if we are given a positive number x and asked what number could be squared in order to give x, there are always two possible answers, one positive, one negative. This is because

(-1)2 = 1, so (-y)2 = (-1)2(y)2 = y2

In order to define the square root as a function, we need it to be single valued, so it is conventional to choose the square root of x to be a positive number; however, if you were asked what the square roots (you might also call these the 2nd roots) of x are, with the understanding that we are not talking about the function but the set of numbers that could be squared to give x, you would say there are two. In fact, by the above logic, the same is true for any even numbered root (the 4th root, the 6th root, etc).

When we take the nth root of a complex number, we find there are, in fact, n roots. Clearly this matches what we found in the n = 2 case. In higher n cases, we missed the extra roots because we were only thinking about roots that are real numbers; the other roots of a real number would be complex. First, let's look at the same case as before, the square root of a real number, then we can move on to other cases.

The Complex Plane

For this task, it is most convenient to think of complex numbers in the polar form; that is, instead of writing complex numbers as z = x + iy, or z = (x,y), where we are thinking of the complex plane in Cartesian coordinates, we wish to think of the complex plane in polar coordinates, describing complex numbers as z = Rei θ, where R is the modulus of the complex number and θ is the argument. Both R and θ are real numbers, and R is greater than zero. If this isn't familiar, see the nodes on complex numbers, Euler's formula, and modulus. In the complex plane, the point z = x + iy is represented by the point (x,y), where R is the distance from the origin to that point and theta is the angle that the line from the origin to the point (x,y) makes with the +x axis. It is also important to remember that since θ is an angle, you can add or subtract any integer multiple of 2π and still get back the same angle, so θ gives the same point as θ + 2πk: mathematically:

ei 2π = 1, so ei (θ + 2πk) = ei θ(ei 2π)k = ei θ(1)k = ei θ

It is this fact that leads to the multiple roots of a complex number. Now let’s move on to square roots of a positive real number.

Squaring and Square Roots

In this representation let's define another complex number w = Sei φ, we can see from the normal rules of exponentiation that wn = Sn(ei φ)n = Snei nφ. In order to find the square roots of z, we want to find complex numbers w, where w2 = z. Using what we wrote above for wn with n=2, then we want w2 = S2ei 2φ. Since z = Rei θ, this means that S2 = R and ei 2φ = ei θ. In short, in order for the two points in the complex plane to match up, we need the lengths and the angles to be equal. Notice that squaring w squares the length and doubles the angle. Now in order to find the roots, we just need to solve to find S and φ in terms of R and theta. Ok, so we already said that by our definition S and R are non-negative real numbers (since they're lengths). We already know how to find the positive square root of a positive real number, that's just what the normal square root function does. So, using the normal definition of the square root function for positive real numbers, S=√R. For the angles the relationship ei 2φ = ei θ, is just saying that the angles have to be the same. This could mean 2φ = θ, but we must remember that you also get the same point if you add any integer multiple of 2π to the angle, so actually we can have 2φ = θ + 2πk, where k is an integer. Solving for φ, we get φ = θ/2 + πk. Now of course k could have any integer value, but we only care about the ones that are actually different angles, so we'll only take values of k where we don't get repeats. k=0 gives φ = θ/2. k=1 gives φ = θ/2 + π. k=2 would add 2π, so that (or any larger k) won't give us a new number. We can also take k < 0, like k = -1, which gives φ = θ/2 - π, but if you add 2π to that you just get θ/2 + π, which we already have. So taking k = 0 and k = 1, we get all the solutions, since the other values of k just give us back the same angles plus a full revolution. Now the last important thing is that

ei (θ/2 + π) = ei θ/2ei π = ei θ/2(-1)

So, to sum up, we've found there are two square roots of a complex number; each has S=√R, and they have φ = θ/2 or φ = θ/2 + π, or to put it another way, w = ± √Rei θ/2.

Back to the Real Numbers

Ok, so, if the complex number we started out with z happened to be a positive real number, then the angle with the real axis, θ, is zero. So, z = R. From the formula we just got for the complex roots, the two square roots would be z = ± √R. Again √R means the positive square root of a positive real number. You may say that was obvious, but it's good that we've shown our method agrees with what we already knew about square roots. Now let's look at a negative real number, where the angle with the real axis is θ = π, so z = Rei π = -R. Using the our formula for the complex roots, we find the two roots are w = √Rei π/2 = i√R and w = √Rei (π/2 +π) = -i√R. If we choose specifically z = -1, then R =1 and we see the roots are w = ± i, which also confirms what we know about i. So, it seems things make sense for the square root. Now let's look at the nth root.

The nth Root

We're finding all the roots w of the number z, such that wn =z. wn = Snei nφ as we said before and z = Rei θ. Again, for the complex numbers to be equal we need the lengths to be the same and the angles to be the same or differ by an integer number of complete rotations by 2π. So S = R1/n (the nth root) as it is defined on positive real numbers, and nφ = θ + 2πk.. Dividing the second equation through by n, we have φ = θ/n + 2π(k/n). So which values of k do we keep? If we keep putting in higher values for k, eventually we will have added 2π and come back to a number we already have, because we've gone around a complete rotation in the complex plane. Looking at the equation for φ, we see that if k < n then (k/n) < 1, so the angle we're adding 2π(k/n) < 2π, so those values are all different. When k = n we'd get back to the same value as k=0. k < 0 also isn't useful, because k = -1 gives the same point as k = n-1¹:

ei 2π(-1/n) = ei (2π(-1/n) + 2π) = ei (2π(-1/n) + 2πn/n) = ei 2π(n-1)/n.

This is because going clockwise by 2π/n is the same as going counter-clockwise by 2π - 2π/n. If we now count up our roots, we have different roots for k = 0, 1, ..., n-1. All together that makes n roots. They have the form

w = R1/nei (θ/n + 2πk/n) for k = 0, 1, ..., n-1

Other Remarks

I just wanted to go back briefly to the idea of the nth complex root as a function. You can define a function that gives you one of the complex roots, but you have to make a decision as to which one. This is often done by deciding on a branch cut, which is a topic for another node.

¹This is related to the idea of modular arithmatic, since we are really working in a system modulo 2π.

Please let me know if you find something unclear, even if you think it's just your problem understanding, because it may well be that the presentation is not actually clear. I also struggled, as usual, with how to format the math to make things look the most clear. Mainly, I'm not sure if it's clear what is in the exponents and what is not. Please let me know.

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