Complex Numbers - Roots of Unity Question (1 Viewer)

[McLake]

Question:

i) Find roots of z^5 + 1 = 0
ii) Factor z^5 + 1
iii) Deduce cos(pi/5) + cos(3pi/5) = 1/2
and cos(pi/5)*cos(3pi/5) = -1/4
iv) <Maybe Later>

Answer:

i) Easyish: 1, +/- cos(pi/5), +/- cos(3pi/5)

ii)Little Harder: z^5 +1 = (z + 1)(z^2 - 2zcos(pi/5) + 1)(z^2 - 2zcos(3pi/5) + 1)

iii) Expand from ii)
z^5 - 1 = (z + 1)(z^4 - 2z^3cos(3pi/5) + z^2 - 2z^3cos(pi/5) + 4z^2cos(pi/5)cos(3pi/5) - 2zcos(pi/5) +z^2 - 2zcos(3pi/5) + 1 )

z^5 - 1 = (z^5 - 2z^4cos(3pi/5) + 2z^3 - 2z^4cos(pi/5) + 4z^3cos(pi/5)cos(3pi/5) - 2z^2cos(pi/5) - 2z^2cos(3pi/5) + z)
+ (z^4 - 2z^3cos(3pi/5) + z^2 - 2z^3cos(pi/5) + 4z^2cos(pi/5)cos(3pi/5) - 2zcos(pi/5) +z^2 - 2zcos(3pi/5) + 1)

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

(equate coefficients of z):
0 = 1 - 2cos(pi/5) - 2cos(3pi/5)
cos(pi/5) + cos(3pi/5) = 1/2

(equate coefficients of z^3):
0 = 2 + 4cos(pi/5)*cos(3pi/5) -2cos(pi/5) - 2cos(3pi/5)
cos(pi/5)*cos(3pi/5) = -1/4

 

[freaking_out]

lol, looks like these forums have bcome so inactive, that u post up your own questions wif answers.

 

[Rahul]

using the sketch of a circle of unity helps a great deal and its a perfectly viable way to answer the question.
i think it also explain things very well.

 

[freaking_out]

Originally posted by Rahul
using the sketch of a circle of unity helps a great deal and its a perfectly viable way to answer the question.
i think it also explain things very well.

yeah, thats the method i used to use, until i came across fitzpatrick's formula for working out the roots.

 

[McLake]

Originally posted by freaking_out
lol, looks like these forums have bcome so inactive, that u post up your own questions wif answers.

Someone asked for it, and yes.

 

[Dumbarse]

question
find 3 cube roots unity of

(z-1)^3 = i