Question

Complex number problem attached. Help please!

Answer 1

@kropot72

Answer 2

@phi

Answer 3

\[\alpha+\beta=-p \\ \alpha \beta=q \text{ by vieta's formula }\]

Answer 4

I got the first equation using the quadratic formula. :-)

Answer 5

But I didn't know the second one, thanks!

Answer 6

\[(w+w^2+w^4)+(w^3+w^5+w^6)=-p \\ (w+w^2+w^4)(w^3+w^5+w^6)=q\]

Answer 7

and we need to somehow use w^7 in all of this

Answer 8

w^7=1 *

Answer 9

Right

Answer 10

kinda want to expand the left hand side of that one equation and apply w^7=1 where I can

Answer 11

\[w^{10}+w^{9}+w^8+3w^7+w^6+w^5+w^4=q \\ w^{10}+w^9+w^8+w^6+w^5+w^4+3(1)=q \]

Answer 12

I' m only thinking right now
so don't expect a straight answer right now :p

Answer 13

No worries :-) Take your time

Answer 14

\[w^7=1 \\ w^7-1=0 \\ (w-1)(w^6+w^5+w^4+w^3+w^2+w+1)=0\]

Answer 15

w is not 1

Answer 16

so that other thing has to be 0

Answer 17

\[w^6+w^5+w^4+w^3+w^2+w+1=0 \text{ is the given thing actually }\]

Answer 18

\[(w+w^2+w^4)+(w^3+w^5+w^6)=-p \\ w^{10}+w^9+w^8+w^6+w^5+w^4+3=q\]

so this first equation there can be written as :
\[w^6+w^5+w^4+w^3+w^2+w=-p \\ \text{ add 1 on both sides } \\ w^6+w^5+w^4+w^3+w^2+w+1=-p+1 \\ 0=-p+1\]

Answer 19

so we can find p now

Answer 20

\[w^{10}+w^9+w^8+w^6+w^5+w^4+3=q \\ w^7w^3+w^7w^2+w^7w+w^6+w^5+w^4+3=q \\w^3+w^2+w+w^6+w^5+w^4+3=q \\ w^6+w^5+w^4+w^3+w^2+w+3=q\]
and guess what

Answer 21

we can find q now too

Answer 22

Is w^7-1=(w-1)(w^6+w^5+w^4+w^3+w^2.....)=0 a factoring formula?

Answer 23

\[w^6+w^5+w^4+w^3+w^2+w+1+2=q \\ 0+2=q \]

Answer 24

well it is
but if you don't know how to factor it you can use division to find the long factor there

Answer 25

Right, thanks.

Answer 26

\[x^{n}-1=(x-1)(x^{n-1}+x^{n-2}+x^{n-3}+\cdots +x^3+x^2+x+1)\]

Answer 27

fun problem
i had to think some
I like thinking

Answer 28

That's awesome! Thanks so much!

Answer 29

freckles did a nice job. But it's always nice to see another approach.
As you know , with roots alpha and beta we have
\[ (x-\alpha)(x-\beta) = x^2 -(\alpha+\beta) +\alpha\beta\]
and pattern matching with \( x^2 +px + q \)
\[ p= -(\alpha+\beta) ;\ q= \alpha\beta\]
adding alpha and beta we get
\[ \alpha+\beta = w+w^2+w^3+w^3+w^4+w^5+w^7 =\sum_{n=1}^{n} w^n \\=\left(\sum_{n=0}^{n} w^n\right)-1\]
the sum has the closed form
\[ \sum_{n=0}^{n} w^n= \frac{w^7-1}{w-1} \]
which from the given information, = 0
and thus
\[ \left(\sum_{n=0}^{n} w^n\right)-1=0-1=-1\]
and \( p= -(\alpha+\beta)=1\)

For q, I used freckles idea. Is there another way?