## JoannaBlackwelder one year ago Complex number problem attached. Help please!

1. JoannaBlackwelder

2. JoannaBlackwelder

@kropot72

3. JoannaBlackwelder

@phi

4. freckles

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

5. JoannaBlackwelder

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

6. JoannaBlackwelder

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

7. freckles

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

8. freckles

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

9. freckles

w^7=1 *

10. JoannaBlackwelder

Right

11. freckles

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

12. freckles

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

13. freckles

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

14. JoannaBlackwelder

No worries :-) Take your time

15. freckles

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

16. freckles

w is not 1

17. freckles

so that other thing has to be 0

18. freckles

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

19. freckles

$(w+w^2+w^4)+(w^3+w^5+w^6)=-p \\ w^{10}+w^9+w^8+w^6+w^5+w^4+3=q$ so this first equation there can be written as : $w^6+w^5+w^4+w^3+w^2+w=-p \\ \text{ add 1 on both sides } \\ w^6+w^5+w^4+w^3+w^2+w+1=-p+1 \\ 0=-p+1$

20. freckles

so we can find p now

21. freckles

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

22. freckles

we can find q now too

23. JoannaBlackwelder

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

24. freckles

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

25. freckles

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

26. JoannaBlackwelder

Right, thanks.

27. freckles

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

28. freckles

fun problem i had to think some I like thinking

29. JoannaBlackwelder

That's awesome! Thanks so much!

30. phi

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?