H2 maths complex numbers

quailmaster SG

Find the roots of z^{7 = 1.}

Show that the sum of the roots equal to zero

I managed to find the roots,
they are 1, e2/7 pi , e^ 4/7pi , e^{6/7 pi

e}-2/7 pi , e^ -4/7pi , e^-6/7 pi

How to show that the sum of all the roots equal to zero?

Just use GC?

Thanks in advance

jayh272416 SG

express everything in components

something magical will happen.

wee_ws SG

Hi,

GC cannot be used to show a result. These steps may be useful:

1. Use the fact that e^(θi) + e^(-θi) = 2 cos θ.

2. Use the factor formulae in MF15.

Another question that is easier to show is:

Show that the product of all the roots is 1.

Thanks.

Cheers,
Wen Shih

eagle SG

Another way to sum is to use a GP with first term e^-6/7 pi, common ratio e^2/7 pi and 7 terms

Hence, the sum of all roots =
e^-6/7 pi ( e^2pi - 1) / (e^2/7 pi - 1)

Since e^2pi - 1 = 0, hence the sum of all roots = 0

quailmaster SG

using the trigo method i ended up with

wee_ws SG

Dear eagle,

Your method is most direct and elegant, thanks :)

I realised, upon closer scrutiny, that my method is not feasible.

Cheers,
Wen Shih

frekiwang SG

This is a method that does not require any calculation:

Draw a unit circle, the seventh roots of unity are unit vectors with 2pi/7 in between.

Note that each interior angle of a regular 7-side polygon is 5pi/7

Do some parallel-shift of the 7 unit vectors, you can easily show them to form a closed 7-side polygon. So according to the rule of adding vectors, the sum of these 7 unit vectors is 0.

wee_ws SG

Hi frekiwang,

This geometrical method is elegant, thanks for sharing!

Cheers,
Wen Shih

wee_ws SG

Hi,

Here is a diagram to clarify frekiwang's geometrical approach:

http://wenshih.files.wordpress.com/2010/02/sum-of-roots-explanation1.pdf

Since the roots sum to 1, we can show that

cos (2pi/7) + cos (4pi/7) + cos (6pi/7) = -1/2

Thanks.

Cheers,
Wen Shih