Cube Roots of Unity HELP! (1 Viewer)

[megan09]

HI I’m in desperate need of help!
I can’t seem to understand the concept of cube roots of unity. Can someone please show me how to attempt the following question:
If Z is one of the three cube roots of unity, find the two possible values of z^2 + z + 1 .
I tried using the formula:
Z = nthroot(r) [cos(t/n + 2kPi/n) + isin(t/n + 2kPi/n)] and found the three roots and substituted each one into z^2 + z + 1. BUT firstly I got 3 answers when they wanted 2 and none of them matched the answers in the back of the book!
What should I do??
Any help would be appreciated

 

[shuning]

formula? there isnt really a formula....

z^3 = 1
z^3 - 1 = 0
(z-1)(z^2+z+1)=0
therefore (z^2+z+1) = 0
is this 1 of the answers?

 

[fuckit1991]

going on from shunning's idea, why don't you use the quadratic formula to get the two values since now you know it is equal to 0. You'll get both complex roots.

z=(-1 (plus minus)iroot3) over 2

 

[shuning]

normally in a question like this, it should say z is a complex root of unity therefore when we have (z-1)(z^2+z+1)=0 we are sure it's (z^2+z+1)=0 not (z-1)=0

maybe there are other ways to do this question

 

[megan09]

hi thanks for replying so fast!
i tried both of these methods, i understand equating it to 0 but i'm not sure of where they are getting 3

 

[megan09]

i found the exact same question posted on yahoo, take a look at answer 2, now that i see the setting out i think i may just understand it!
A complex number roots question.? - Yahoo! Answers

 

[roadrage75]

formula? there isnt really a formula....

z^3 = 1
z^3 - 1 = 0
(z-1)(z^2+z+1)=0
therefore (z^2+z+1) = 0
is this 1 of the answers?

i'd say so...

notice there're not asking you to find out all the cube roots of unity. They're asking you to find the possible values for

z^2 + z + 1, given that z is a cube root of unity

as shuning said, z^3 = 1. therefore, (z-1)(z^2 + z + 1) = 0

so either z = 1 --> z^2 + z + 1 = 3

or z^2 + z + 1 = 0 (who cares what z is, the question doesnt ask for this)

therefore, z^2 + z + 1 = 3 or 0.

 

[shuning]

just wondering, i understand y it equals to 0, but why 3?

 

[L]

z = w

 

[roadrage75]

just wondering, i understand y it equals to 0, but why 3?

can you understand how I got to this step:

(z-1)(z^2 +z + 1) = 0


now z = 1 satifies this equation, right? (this makes sense as 1 is one of the three cubic roots of 1)

if z = 1, z^2 + z + 1 = 3 (by mere substitution)

the only other way (z-1)(z^2 +z + 1) can = 0 is if z^2 + z + 1 = 0 itself

 

[shuning]

can you understand how I got to this step:

(z-1)(z^2 +z + 1) = 0


now z = 1 satifies this equation, right? (this makes sense as 1 is one of the three cubic roots of 1)

if z = 1, z^2 + z + 1 = 3 (by mere substitution)

the only other way (z-1)(z^2 +z + 1) can = 0 is if z^2 + z + 1 = 0 itself

oh thanks, understand now