How to find the n-th roots of a complex number?

To begin, we recall a fact about complex numbers that we saw last time: any complex number, written in the form a + bi, can be represented as a + bi = rexi = rcos(x) + ri sin(x), using the Euler's formula. Where the value of r and x are just the radius and the angle in polar coordinate, on the Cartesian plane, which in this case represents the complex plane where a complex number a + bi is represented by the point (a,b). To be more precise, r = (a2 + b2)1/2, is the distance between the point (a,b) and the origin, and the angle x can be determined by the signs of a and b, together with the reference angle given by arctan(b/a). The reverse conversion is given by a = rcos(x) and b = rsin(x), as can be easily seen above.

The value r is called the modulus of the complex number, and the angle x is the argument of the complex number. Note that the conjugate of the complex number a + bi, a - bi, has the same modulus but whose argument differs from x by pi.

Just like the usual conversion between points in rectangular coordinate and polar coordinate systems, the value of r is non-negative, and r is zero only if a = b = 0 (and therefore the complex number a + bi is 0 itself). The angle x, by convention, is restricted between 0 and 2pi. This way, any point (r,x) can be mapped uniquely to a point (a,b), and vice versa, except when a = b = 0 (when r = 0, and x can be unspecified).

With the above in mind, we just need to recall one more bit of prior knowledge on laws of exponential functions in order to derive the general method of extracting the n-th roots of an arbitrary complex number. The specific rule in question is:

(c1ea)(c2eb) = (c1c2)ea+b.

That is, when multiplying two (or more) exponential terms, multiply their coefficients and add their exponents.

Now we can try some "reverse engineering" to find the formula for finding the n-th roots of a complex number. Suppose another complex number c + di = r1ezi is one of the n-th root of a + bi = rexi. Then (r1ezi)n = (r1)nenzi = rexi. Compare the coefficients and exponents, we see that (r1)n = r and nz = x. So r1 = r1/n, and z = x/n. Well, not quite - the modulus r1 is fine, but there are more to it in determining the argument z. Remember that the argument is restricted between 0 and 2pi, i.e., if the exponent nz is outside of this range then it is modified accordingly in the usual way. Therefore, an angle such as x/n + 2pi/n, when multiplied by n, will also give us the same angle x. In fact, x/n, x/n + 2pi/n, x/n + 4pi/n, ... , x/n + 2kpi/n, where k runs from 1 to n-1, are all angles between 0 and 2pi that give the same angle x when multiplied by n. (That is, after being multiplied by n they all differ from x by some multiple of 2pi.) Notice that there are n such angles between 0 and 2pi. In set notation, this set can be written as {x + 2kpi : k = 0, 1, ... , n-1}.

Therefore, the n-th roots of any given complex number a + bi is:

r1/n e((x + 2kpi)/n) = r1/n [cos ((x + 2kpi)/n) + i sin ((x + 2kpi)/n)],

where r = (a2 + b2)1/2 is the modulus of the given complex number, r1/n being the positive real n-th root of r (such a root always exists since r is always a non-negative real number), x is the argument of the given complex number, and k = 0, 1, 2, ... , n-1.


Another word, the n-th roots (there are n of them) of a complex number a + bi is another complex number c + di. Where the real part is c = r1/n cos ((x + 2kpi)/n), and its imaginary part is given by d = r1/n sin ((x + 2kpi)/n), k = 0, 1, ... , n-1.

A side note: Algebraically, the n-th roots of a + bi are just the roots of the (complex) polynomial xn - (a + bi). It is a degree n polynomial, so there are n such roots, according to the Fundamental Theorem of Algebra.

If we take a + bi = 1 (= 1 + 0i, i.e., a = 1 and b = 0) and any positive integer n, then we get a special family of complex numbers. Those are the n-th roots of unity (or the n-th roots of 1). For a specific value of n, the n-th roots of unity are the n distinct roots of the polynomial xn - 1. In this case, the modulus r is 1, and the argument x is 0. So the roots are just cos (2kpi/n) + i sin (2kpi/n), k= 0, 1, ... , n-1. When k = 0, the corresponding root is of course 1, which is always a n-th root of itself, for any value of n. Roots of unity are of particular interest, which we will further explore next time.

Next time: Roots of Unity