The set of n^{th} roots of a complex number z contains all complex numbers w such that
w^{n} = z. If we use the polar notation for complex numbers, so that z =
Re^{i θ} and w = Se^{i φ}, then we find there are n roots of the form:

S = R^{1/n} and φ = θ/n + 2π(k/n) where k = 0, 1, ..., n-1

## Explanation

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} = y^{2}

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 2^{nd} 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 4^{th} root, the 6^{th} root, etc).

When we take the n^{th} 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 = Re^{i 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:

e^{i 2π} = 1, so e^{i (θ + 2πk)} =
e^{i θ}(e^{i 2π})^{k} = e^{i θ}(1)^{k} = e^{i 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 = Se^{i φ}, we can see
from the normal rules of exponentiation that w^{n} = S^{n}(e^{i φ})^{n} = S^{n}e^{i nφ}.
In order to find the square roots of z, we want to find complex numbers w, where w^{2} =
z. Using what we wrote above for w^{n} with n=2, then we want w^{2} = S^{2}e^{i 2φ}.
Since z = Re^{i θ}, this means that S^{2} = R and e^{i 2φ} = e^{i 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 e^{i 2φ} = e^{i 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

e^{i (θ/2 + π)} = e^{i θ/2}e^{i π} = e^{i 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 = ±
√Re^{i 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 = Re^{i π} = -R. Using the our formula for the complex
roots, we find the two roots are w = √Re^{i π/2} = i√R and w =
√Re^{i (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 n^{th} root.

### The n^{th} Root

We're finding all the roots w of the number z, such that w^{n} =z. w^{n} = S^{n}e^{i nφ}
as we said before and z = Re^{i 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 = R^{1/n} (the n^{th} 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:

e^{i 2π(-1/n)} = e^{i (2π(-1/n) + 2π)} = e^{i (2π(-1/n) + 2πn/n)} =
e^{i 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 = R^{1/n}e^{i (θ/n + 2πk/n)} for k = 0, 1, ..., n-1

### Other Remarks

I just wanted to go back briefly to the idea of the n^{th} 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.

¹

This 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.