Tuesday, March 12, 2013

Taking Complex Roots (1)

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 earlier 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 + isin q)2 = cos2q + isin 2q

Expand the left side:

(cos q + isinq)2 = (cosq + isinq)(cosq + isinq)

= cos2(q) + 2icos(q)sin(q) – sin2(q)

We now gather the real and imaginary parts:

cos(2q) = cos2(2q) – sin2(2q)

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


Now, 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……wn 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…..

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 (q = 0 or 2p) and with the angles given at the boundaries of the n segments
See, e.g. http://brane-space.blogspot.com/2011/02/isomorphism-between-galois-groups-and.html, for the case of taking the fifth root and the locus of points on the unit circle)   With reference to the diagram in the above link, the reader should be able to sketch  the diagram giving the results for n = 3, .i.e. for the 3rd 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 w1,2 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)

Problem for the Math Maven:

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

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