## Saturday, March 20, 2010

### Back to Complex Numbers: Roots

It’s time to return to the fascinating realm of the complex numbers, now to show how to take nth roots, In earlier segments we saw how any complex number, z, can be written in polar form, e.g.:

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

Now think of this:

If n is a positive integer we may express z^n:

z^n = z* z* z* z …….z (n factors)

But if: z= rcis(theta) then

z^n = (r cis(theta))^n = r^n cis(theta + theta + theta +………theta) (e.g. n summands)

and z^n = r^n cis (n(theta))

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

Now let r=1 in the preceding, so:

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

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 theta) and sin (n theta) as polynomials of degree n in cos (theta) and sin(theta).

For example, let n =2 in the above eqn.- and for sake of ease in writing, let theta = q

(cos q + isinq)^2 = cos2q + isin 2q

Expand the left side:

(cos q + isinq)^2 = (cosq + isinq)(cosq + isinq) = cos^2(q) + 2icos(q)sin(q) – sin^2(q)

We now gather the real and imaginary parts:

cos(2q) = cos^2(2q) – sin^2(2q)

sin(2q) = i2cos(q)sin(q)

Roots: Now, if z=r cis(theta) is a complex number different from 0, and n is a positie 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 = rho cis(a) be an nth root of z = rcis(theta) so w^n = z

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

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

Problem: take the 3rd roots of 1.

Here: r = 1

The angle is easily determined using and applying the geometry from Galois extensions, for which the number of roots n, divides a unit circle into n equal segments – starting at r = 1 (theta = 0 or 2 pi) and with the angles given at the boundaries of the n segments.

For example the diagram shown give the results for n = 3. This yields:

pi/3 = 120 deg

2 pi/ 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(120) +isin(120)) = - ½ +/- i(0.866)
For the 3rd root of unity: (k =+/-2)

w2 = [1] (cos(240) + isin(240) = - ½ -/+ i(0.866)

Note that the +/- signs for the roots w2, and w3 yield two redundant roots. Eliminating them (e.g. removing the secondary sign root duplicates in each case) we have:

w0 = 1

w1 = - ½ + i(0.866)

w2 = - ½ - i(0.866)

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