Complex No. Q. (1 Viewer)

[Sparcod]

I should be shot for asking such an easy question but here goes.


Factorise z^5 -1 into real linear and quadratic factors.

Hence deduce:
1.cos 2n/5 +cos 4n/5 = -1/2
2.cos 2n/5 x cos 4n/5 =-1/4

sorry but note: n is pi. (that's the closest thing to pi)

Thanks to all those who help.
Also, whilst doing it, time yourself to find out your working time. (not the time spent writing). Thanks. Just curious to know how time-consuming this is.

 

[pLuvia]

z⁵=1
z⁵= cis 0
z= cis 2kn/5 (where k is 0,+1,+2
Subbing the k's in

z = 1, cis2n/5, cis-2n/5, cis4n/5, cis-4n/5

z⁵-1 = (z-1)(z⁴+z³+z²+z+1)

= (z-1)(z-cis2n/5)(z-cis-2n/5)(z-cis4n/5)(z-cis-4n/5)

Note: (z-cis2n/5)(z-cis-2n/5)
= (z-cos2n/5+sin2n/5)(z-cos2n/5-isin2n/5)

Simplify by expansion

= z²-2cos 2n/5z + 1

Similarly with (z-cis4n/5)(z-cis-4n/5)

= z²-2cos4n/5z + 1

.: (z-1)(z²-2cos 2n/5z + 1)(z²-2cos4n/5z + 1)

I'm not sure if this is what you want

Edit: Not sure how to do the next bit and didn't bother timing but it was long

 

[Sparcod]

z^5 -1=0
z^5=(z-1)(z4+z3+z2+z+1)=0
then divide by z+1
z^4+z^3+z^2+z+1= (z^2-2z cos 2n/5 +1) (z^2+ 2z cos 4n/5+1)

ok..then what??

 

[pLuvia]

How much more do you want lol? I don't know what to do from there

 

[Raginsheep]

Use sums and product of the non-real roots.

 

[Sober]

z^5 -1=0
z^5=(z-1)(z4+z3+z2+z+1)=0
then divide by z+1
z^4+z^3+z^2+z+1= (z^2-2z cos 2n/5 +1) (z^2+ 2z cos 4n/5+1)

ok..then what??

Expand the right hand side then group the powers of z and equate each individually with the coefficients of z on the left and you will find your identities.