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

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freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\alpha+\beta=p \\ \alpha \beta=q \text{ by vieta's formula }\]

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1I got the first equation using the quadratic formula. :)

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1But I didn't know the second one, thanks!

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[(w+w^2+w^4)+(w^3+w^5+w^6)=p \\ (w+w^2+w^4)(w^3+w^5+w^6)=q\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3and we need to somehow use w^7 in all of this

freckles
 one year ago
Best ResponseYou've already chosen the best response.3kinda want to expand the left hand side of that one equation and apply w^7=1 where I can

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[w^{10}+w^{9}+w^8+3w^7+w^6+w^5+w^4=q \\ w^{10}+w^9+w^8+w^6+w^5+w^4+3(1)=q \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I' m only thinking right now so don't expect a straight answer right now :p

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1No worries :) Take your time

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[w^7=1 \\ w^71=0 \\ (w1)(w^6+w^5+w^4+w^3+w^2+w+1)=0\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so that other thing has to be 0

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[w^6+w^5+w^4+w^3+w^2+w+1=0 \text{ is the given thing actually }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[(w+w^2+w^4)+(w^3+w^5+w^6)=p \\ w^{10}+w^9+w^8+w^6+w^5+w^4+3=q\] so this first equation there can be written as : \[w^6+w^5+w^4+w^3+w^2+w=p \\ \text{ add 1 on both sides } \\ w^6+w^5+w^4+w^3+w^2+w+1=p+1 \\ 0=p+1\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[w^{10}+w^9+w^8+w^6+w^5+w^4+3=q \\ w^7w^3+w^7w^2+w^7w+w^6+w^5+w^4+3=q \\w^3+w^2+w+w^6+w^5+w^4+3=q \\ w^6+w^5+w^4+w^3+w^2+w+3=q\] and guess what

freckles
 one year ago
Best ResponseYou've already chosen the best response.3we can find q now too

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1Is w^71=(w1)(w^6+w^5+w^4+w^3+w^2.....)=0 a factoring formula?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[w^6+w^5+w^4+w^3+w^2+w+1+2=q \\ 0+2=q \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3well it is but if you don't know how to factor it you can use division to find the long factor there

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1Right, thanks.

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[x^{n}1=(x1)(x^{n1}+x^{n2}+x^{n3}+\cdots +x^3+x^2+x+1)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3fun problem i had to think some I like thinking

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.1That's awesome! Thanks so much!

phi
 one year ago
Best ResponseYou've already chosen the best response.0freckles 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 \[ \sum_{n=0}^{n} w^n= \frac{w^71}{w1} \] which from the given information, = 0 and thus \[ \left(\sum_{n=0}^{n} w^n\right)1=01=1\] and \( p= (\alpha+\beta)=1\) For q, I used freckles idea. Is there another way?
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