need help! complex numbers (1 Viewer)

[blackfriday]

1. form the equation whose roots are 2+w and 2+w^2, where w is a complex root of unity.

2. if p is one of the complex roots of x^7-1=0, form the equation whose roots are: p+p^2+p^4 and p^6+p^5+p^3.

help would be appreciated.

thanks

 

[Rorix]

use sum and product

 

[Estel]

x^2+ax+b=0 (2 roots)
-a = 4 + w + w ^2 = 3
b = (2+w)(2+w^2) = 4 + 2w + 2w^2 + w^3 = 3
:. required eq is x^2-3x+3=0

x^2+ax+b=0 (2 roots)
-a = p+p^2...+p^6
p, p^2,...p^7 are the seventh roots of unity
p^7 = 1
-a = -1, a = 1
b= (p+p^2+p^4)(p^6+p^5+p^3) = p^7 + p^6 + p^4 + p^8+ p^7 + p^5 + p^10 + p^9 + p^7
= 2 + p^7 + p^6 + p^4 + p^5+ p^3 + p^2 + p =2
:. req eq is x^2+x+2=0

gees you're movin ahead in 4U awfully fast.

 

[blackfriday]

*smacks head on wall*

so...a+b=-b/a and ab=c/a right?

so... 4+w+w^2=-b/a and
4+2w^2+2w+w^3=c/a

...

 

[blackfriday]

thanks for the help so far. i think we're moving ahead too fast for my brain to process. i did ext 2 maths purely to get out of my maths class because my teacher sucked and i wasnt a complete lost case at maths. but complex numbers has disproved that.

just a question...

for the first one: 4 + w + w ^2 = 3...why does it equal 3? and same question for the next one.

 

[Estel]

4+w+w^2 = 3 + (1+w+w^2) = 3

[z^3 = 1
roots are 1 + w + w^2
sum of roots =-b/a = 0]

lol we just got up to modulus.

 

[blackfriday]

well you're obviously amazingly intelligent. i get it now.

now im trying to work out the next one.

 

[lfc_reds2003]

Are u in yr 11 going into 12

are u kids in yr 11 going in 12 cos if so we havenyt started ext 2 yet.... hope i am not the only one!

 

[Estel]

If we were in Yr 12, would be spending such inordinate amounts of time online?

 

[Slidey]

We are about to start complex roots of unity.

It sucks though; I keep making silly errors - I am rushing things, which is perhaps quite bad.

 

[Rorix]

If we were in Yr 12, would be spending such inordinate amounts of time online?

..............im in yr12

 

[Jase]

So am i. and theres about three more days to go until the day i may never have to find the complex nth roots of unity of 1 ever again! woohoo! But until then... aww.

 

[Supra]

rnt u gonna do a lil bit of maths at uni?

 

[blackfriday]

just a question from before...

how did you know the general equation was x^2+ax+b? how do you know if there a constant if front of the x^2 term?

 

[Shuter]

I think (and I'm not sure but I really should be seeing as the HSC is 3 days away) that he just was lazy and did it in the form:

x^2 + ax + b

in reality it should be written as:

a(x^2) + bx + c

which which case you simply divide by a if there's a co-efficient out the front.

 

[Estel]

ax^2+bc+c=0 is the general form

however, we can write that as say x^2+dx+e=0 without loss of generality as you can divide any quadratic by it's leading coefficient.

Note to Yr12's : stop proving me wrong and get back studyin

 

[blackfriday]

sorry i meant how do you find out the co-efficient of the leading term?

 

[Rorix]

its not needed
e.g.
2x^2 + 4x + 1 = 0
x^2 + 2x + 1/2=0