Need help to solve this equation "z^4+z^2+1=0" ?...

-Syn- · 07-08-2008 7:03pm #1

Studying for a repeat exam in two weeks and cant get equations like this, think its something to do with de moivres theorem..

Cokehead Mother · 07-08-2008 7:57pm

I can think of one way of doing this but it doesn't involve DeMoivre's and it isn't very nice. Maybe someone will be able to offer a quicker method?

Michael Collins · 08-08-2008 2:37pm

A slightly quicker way:

When you get down to the z^2, using Cokehead's method, then use de Movire's theorem. i.e.

z^2 = (-1 ± i root(3))/2 = cos(2pi/3) ± i sin(2pi/3)

so z = ± (cos(pi/3) ± sin(pi/3)) = ± (1 ± i root(3))/2

which gives four answers, two pairs of complex conjugates - as expected for a fourth order polynomial with real coefficients.

I suspect there's a simpler way to do it using the properties of the roots of unity, but this should do for now.

MathsManiac · 08-08-2008 6:17pm

Here's a way that would require you to have the ability to recognise certain special forms of complex polynomials:

You might notice that multiplying your given equation by z^2 - 1 would give you z^6 - 1.

The solutions of this latter equation are the "sixth roots of unity", recognisable to those in the know as (in degrees): CiS(360n/6), which is CiS(60n), for n=0,1,2,3,4,5. [CiS(X) is shorthand for Cos(X) + iSin(X).]

The step of multiplying by the z^2 - 1 introduced the two roots 1 and -1, so the remaining four are the roots of the original equation:
1/2 + (1/2)*i*sqrt(3), 1/2 - (1/2)*i*sqrt(3), -1/2 + (1/2)*i*sqrt(3), -1/2 - (1/2)*i*sqrt(3).

And, by the way, how did I think of doing the first step? By knowing that z^n + z^(n-1) + z^(n-2) + ... + z + 1 is [z^(n+1) - 1] / (z - 1), which is a well known result in this topic and easily proved. This looks like that, but with z^2 for z.

MathsManiac · 08-08-2008 6:18pm

...and forgot to mention that it's DeMoivre that gives you the sixth roots of unity.

MathsManiac · 08-08-2008 6:52pm

And just thought of another method, although I'm not sure I'd have spotted this if I didn't already know the solutions: the original equation factorises as (z^2 + z + 1)*(z^2 - z +1), and you could just solve these two quadratics with the formula or by completing the square. (But there's not a sniff of DeMoivre this way.)