A complex number trig proof for you all...

[Weyoun]

Righto.

We're studying factorisation over the complex field, and have to give trig proofs for z^n + 1 = 0 where n is an even number. Generally, what I have done in questions preceding this one is, because there were z^n - 1 = 0 and n was odd, divide z^n - 1 by z - 1, resulting in, for example:

z^5 - 1 = 0
(z^5 - 1)/(z - 1) = z^4 + z^3 + z^2 + z + 1

and equated the coefficients of the different powers of z using the factors I had earlier obtained.

However.

I stumbled upon this question, paired with another, slightly easier one. Here goes...

Find the roots of z^6 + 1 = 0, and hence resolve z^6 + 1 = 0 into real quadratic factors; deduce that
cos3Theta = 4(cosTheta - cos(PI/6))(cosTheta - cos(PI/2))(cosTheta - cos(5PI/6))

I have absolutely no idea how to attack this problem, due to the introduction of the open ended Theta variable. I found the quadratic factors of z^6 + 1, but how would I get this trig identity exactly? If someone could please explain a way of deducing this trig identity instead of simply simplifying the right hand side to equate cos3Theta or appropriate substitutions? I'm really in quite a bind here, our teacher has no notes on them, and to be honest, hasn't shown any signs that he knows how to solve it either. If someone with an education superior to that of Australian Public schools could explain this to me, it would be much appreciated
Andrew

 

[m0ti]

Well, it's 2 AM, and I I'm rather tired, but here's what I think. I'm not really sure if it's trig or not (it probably is, but you could, perhaps, use other things to solve it - like suitable transforms). Anyways, it appears to me that they've introduced a new variable y = exp(Theta), and you're looking at a calculation for Re(y^3).

I'm not sure how they get at the lhs (left-hand side), but they seem to be only using the factors from the "positive" side of the unit circle... like (z^2)^3 + 1 = 0?

Not sure if this helps any. Like I said, it's 2 AM.

Hopefully it turns on a light bulb or something.

 

[Shalmanese]

cos 5pi/6 = -cos pi/6 so you get a difference of 2 squares

cos 3theta = 4(cos^2 theta - cos^2 pi/6)(cos theta - cos pi/2)
cos pi/2 = 0 so we get 4(cos^2 theta - cos^2 pi/6)(cos theta) which is 4(cos^3 theta - 3/4cos theta)
expanding, we get 4 cos^3 theta - 3 cos theta

graphamatic tells me I am right so far but I still need to convert it into cos 3 theta so:

using addition angle formula: cos 3 theta = cos 2theta cos theta - sin 2 theta sin 2theta
using double angle formula: cos 2theta cos theta - sin 2 theta sin 2theta = 2(cos^2 theta -1) cos theta - 2 sin theta cos theta sin theta
simplifies down to: 2 cos^3 theta - cos theta - 2 sin^2 theta cos theta

sin^2 theta = 1-cos^2 theta so it all goes down to 2 cos^3 theta - cos theta - 2cos theta + 2cos^3 theta

which is 4 cos^3 theta - 3 cos theta

QED

 

[m0ti]

THAT"S not called a deduction.

That's just simplifying the rhs, which is exactly what he didn't ask for!

 

[Shalmanese]

Sorry, my bad.

I am a bit puzzled at it asking specifically to deduce it though. AFAIK, Australian pulic schools arent meant to be taught deduction, only a bit of induction in Yr12 so I dont know why they would be asking specifically to deduce it.

 

[Mday]

this is such trivial work btw...

 

[Bluga]

in Engineering Math:

roots of z^6 + 1 = 0 means we want to find the 6th roots of -1.

|-1| = 1

and an argument of -1 is 0, therefore the 6th roots of -1 are

(1)exp((0 + 2K(pi))i/6), where k = 0, 1, 2, 3, 4, 5

 

[Agent004]

<< in Engineering Math:

roots of z^6 + 1 = 0 means we want to find the 6th roots of -1 >>



Partly true, but it said find 'roots' meaning all 6 roots.

Generally, equatiosn are in the form of Z^n = a

Rearranging the equation gives you Z^6 = -1, abs of Z is 1, a is -1.

Now imagine you have the complex plane, you have a circle centred at 0,0 and radius 1. All the solutions/roots lies on this circle and they are equally spaced apart.

For the root, you need to know the arg z/theta of each roots


theta = arg a / n

Where n is the nth power of z

Since arg (-1) = pi

Theta = pi/6

Any root may be expressed as abs z (cos theta + i sin theta)

1st root is 1 (cos pi/6 + i sin pi/6), you can work this out it's real number answer

The subsequence roots may be obtain by adding 2pi/n to the existing angle

The second root is 1 (cos 3pi/6 + i sin 3pi/6 ) amd etc

Hope this helps

Edit: If you work out all the roots of this equations, the deduction just drops out