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JoannaBlackwelder

  • one year ago

Complex number problem attached. Help please!

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  1. JoannaBlackwelder
    • one year ago
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  2. JoannaBlackwelder
    • one year ago
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    @kropot72

  3. JoannaBlackwelder
    • one year ago
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    @phi

  4. freckles
    • one year ago
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    \[\alpha+\beta=-p \\ \alpha \beta=q \text{ by vieta's formula }\]

  5. JoannaBlackwelder
    • one year ago
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    I got the first equation using the quadratic formula. :-)

  6. JoannaBlackwelder
    • one year ago
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    But I didn't know the second one, thanks!

  7. freckles
    • one year ago
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    \[(w+w^2+w^4)+(w^3+w^5+w^6)=-p \\ (w+w^2+w^4)(w^3+w^5+w^6)=q\]

  8. freckles
    • one year ago
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    and we need to somehow use w^7 in all of this

  9. freckles
    • one year ago
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    w^7=1 *

  10. JoannaBlackwelder
    • one year ago
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    Right

  11. freckles
    • one year ago
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    kinda want to expand the left hand side of that one equation and apply w^7=1 where I can

  12. freckles
    • one year ago
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    \[w^{10}+w^{9}+w^8+3w^7+w^6+w^5+w^4=q \\ w^{10}+w^9+w^8+w^6+w^5+w^4+3(1)=q \]

  13. freckles
    • one year ago
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    I' m only thinking right now so don't expect a straight answer right now :p

  14. JoannaBlackwelder
    • one year ago
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    No worries :-) Take your time

  15. freckles
    • one year ago
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    \[w^7=1 \\ w^7-1=0 \\ (w-1)(w^6+w^5+w^4+w^3+w^2+w+1)=0\]

  16. freckles
    • one year ago
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    w is not 1

  17. freckles
    • one year ago
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    so that other thing has to be 0

  18. freckles
    • one year ago
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    \[w^6+w^5+w^4+w^3+w^2+w+1=0 \text{ is the given thing actually }\]

  19. freckles
    • one year ago
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    \[(w+w^2+w^4)+(w^3+w^5+w^6)=-p \\ w^{10}+w^9+w^8+w^6+w^5+w^4+3=q\] 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
    • one year ago
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    so we can find p now

  21. freckles
    • one year ago
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    \[w^{10}+w^9+w^8+w^6+w^5+w^4+3=q \\ w^7w^3+w^7w^2+w^7w+w^6+w^5+w^4+3=q \\w^3+w^2+w+w^6+w^5+w^4+3=q \\ w^6+w^5+w^4+w^3+w^2+w+3=q\] and guess what

  22. freckles
    • one year ago
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    we can find q now too

  23. JoannaBlackwelder
    • one year ago
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    Is w^7-1=(w-1)(w^6+w^5+w^4+w^3+w^2.....)=0 a factoring formula?

  24. freckles
    • one year ago
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    \[w^6+w^5+w^4+w^3+w^2+w+1+2=q \\ 0+2=q \]

  25. freckles
    • one year ago
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    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
    • one year ago
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    Right, thanks.

  27. freckles
    • one year ago
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    \[x^{n}-1=(x-1)(x^{n-1}+x^{n-2}+x^{n-3}+\cdots +x^3+x^2+x+1)\]

  28. freckles
    • one year ago
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    fun problem i had to think some I like thinking

  29. JoannaBlackwelder
    • one year ago
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    That's awesome! Thanks so much!

  30. phi
    • one year ago
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    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?

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