Question (1 Viewer)

[ProGT408]

a) fully factorise z^6-1=0 over the rational field
b) solve z^2+z^4+z^6 over the complex field

how do you use part a to do part b?

 

[synthesisFR]

a) fully factorise z^6-1=0 over the rational field
b) solve z^2+z^4+z^6 over the complex field

how do you use part a to do part b?

Cuz for Part a u do (z^2)^3 - 1 = 0
Then use the difference of two cubes formula thing ye
For part b factorise z^2 out to get 1 + z^2 + z^4 which is in the bracket of the part a one mad ting innit
Then use the solutions from a to get the ones in part two by doing normal roots of unit minus the solution z=1 cus that’s the other bracket cuh
Then u basically get the rest of the solutions fam
Free the mandem while your at it

 

[liamkk112]

a) fully factorise z^6-1=0 over the rational field
b) solve z^2+z^4+z^6 over the complex field

how do you use part a to do part b?

for a) z^6-1 = 0 -> (z^3-1)(z^3+1)=0 -> (z-1)(z^2+z+1)(z+1)(z^2-z+1)=0 using difference of squares, now z^2+z+1 and z^2-z+1 have complex roots so we do not factorise any further
then for b) using a) to expand some brackets: (z-1)(z+1)(z^4+z^2+1)=0
hence the solutions to z^4+z^2+1 = 0 are the complex 6ths roots of units, namely those with modulus 1 and argument pi/3, 2pi/3, -pi/3, and -2pi/3 (since z = -1, z = 1 are the solutions to z-1 = 0, z+1 = 0, meaning z^4+z^2+1 must have the other solutions to z^6 = 1)

now z^2+z^4+z^6 = z^2 (z^4 + z^2 +1) hence the solutions are z = 0 and the complex 6th roots of unity above

 

[ivanradoszyce]

For part a), write
as

a difference of 2 cubes.

Recall that , therefore

The roots where are .

b) Consider

as

In this case we need to find the 6th roots of unity. So let , there for .



Therefore



For the Principle Argument of




Therefore , which means

The values of and , represent the values of and

The values of are the solutions to


Let's now consider



Using the results from part (a)

 

[Luukas.2]

Recognising the GP in the second polynomial allows the link between the two parts to be seen more easily:

​

Henxe, the roots of are the six roots of , with the two roots of , that is, at , removed, and a double root at