Dealing With The Basics Of Complex Numbers (Part 3): Complex Roots

 Just as there are roots of real numbers, for example:

Ö4 = +2  and

Ö3 = +1.732

 So also can one obtain the roots of complex numbers. A term often used is taking the “nth roots” of a complex number (or even a normal number, like 1). For example, in the real number cases above, there are plus and minus valued roots for each of the square roots taken.

In previous instalments we saw how any complex number, z, can be written in polar form, e.g.:

z = r(cos (q) + isin(q))

Now think of this:  If n is a positive integer we may express zⁿ: 

zⁿ = z· z· z· z ……. (n factors)

But if: z= rcis(q) then

zⁿ = (r cis(q))ⁿ = rⁿ cis(q + q + q +………q) (e.g. n summands)

and zⁿ = rⁿ cis (n(q))

In vector language one would say that the length (e.g. radius)    r = êz ê is raised to the nth power and the angle q = arg(z) is multiplied by n.

Now let r=1 in the preceding, so:

cis (n(q)) = (cos (q  + isin(q)) = cos (n q) + isin(n q)

which is De Moivre’s theorem.

Now, if we expand the left side of the preceding equation (by the binomial theorem) and reduce it to standard complex form: z = a + ib, then we obtain formulas for cos (n q) and sin (n q) as polynomials of degree n in cos (q) and sin(q).

For example, let n =2 in the above eqn. then:  (cos q + i sin q)² = cos2q + i sin 2q

Expand the left side:

(cos q + i sinq)² = (cosq + i sinq)(cosq + i sinq)

= cos²(q) + 2 i cos(q)sin(q) – sin²(q)

Then gather the real and imaginary parts:

cos(2q) = cos²(2q) – sin²(2q)

and: we have the identity:   sin(2q) = i2cos(q)sin(q)

If z = r cis(q) is a complex number different from 0, and n is a positive integer, then there are exactly n different complex numbers: w₁, w₂, w₃……w_n each of which is the nth ROOT of z. If we let w = r cis(q) be an nth root of z = rcis(q) so wⁿ = z

It can eventually be shown ( I leave this to readers) that:

(r cis(q))^1/n = (r)^1/n cis(q/n + k (2 p)/n) ,  with k = 0, +1, +2…..


Example Problem: 

Take the 3rd roots of 1.

Here: r = 1

The angle is easily determined using and applying the geometry from Galois extensions, see e.g.

Isomorphism between Galois Groups and triangulated cover groups of form π (X)

for which the number of roots n, divides a unit circle into n equal segments – starting at r = 1 (q = 0 or 2p) and with the angles given at the boundaries of the n segments.
 
With reference to a simple diagram, the reader should be able to sketch the results for n = 3, .i.e. for the 3^rd roots of 1.  This yields:

2p/3 = 120 deg  and  4p/ 3 = 240 deg

For the first root of unity (k =0) so:

w₀ = [1] (cos(0)) = 1

For the 2nd root of unity: (k =+1)

w₁ = [1](cos(2p/3) +isin(2p/3)) = - ½ + i(Ö3/2)

For the 3rd root of unity: (k =+ 2)

w₂ = [1] (cos(4p/ 3) + isin(4p/ 3) = - ½ + i(Ö3/2)

Note that the +/-  signs for the roots w_1,2 yield two redundant roots. Eliminating them (e.g. removing the secondary sign root duplicates in each case) we have:

w₀ = 1

w₁ = - ½ + i((Ö3/2)

w₂ = - ½ - i((Ö3/2)

Suggested Problem:

Find the fourth roots of unity and provide a sketch diagram to locate them.