Complex Number (1 Viewer)

[lollol]

let P=cos(2pi/7)+isin(2pi/7). the complex number a=P+P^2+P^4 is a root of the quadratic equation x^2+ax+b=0, where a and b are real.

1. prove 1+P+P^2+.....P^6=0

2. find the values of the coefficients a and b

3. deduce that -sin(pi/7)+sin(2pi/7)+sin(3pi/7)=root 7 times 0.5

 

[Yip]

1. P=cos(2pi/7)+isin(2pi/7)
P^7=cis(2pi)=1
P^7-1=0
(P-1)(1+P+P^2+...+P^6)=0
P is complex, so P=/=1
1+P+P^2+...+P^6=0
2. P+P^2+P^4 is a root of the quadratic x^2+ax+b=0, whose coefficients are real
Thus its conjugate, P^-1 + P^-2 + P^-4 is also a root
P^7=1
P^-1=(P^-1)(1)=P^-1(P^7)=P^6
Likewise, P^-2=P^5, P^-4=P^3
Sum of roots: P+P^2+P^4+P^6+P^5+P^3=-a
From 1., -a=-1
a=1
Product of roots: (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
=3+P+P^2+...+P^6 (since P^7=1)
=b
b=2
3. Quadratic is x^2+x+2
x=[-1+-(root7)i]/2=P+P^2+P^4,P^6+P^5+P^3
Im[z]=+-[root7]/2=Im[P+P^2+P^4], Im[P^6+P^5+P^3]
Im[P^6+P^5+P^3]=sin(12pi/7)+sin(10pi/7)+sin(6pi/7)<0
Thus, Im[P+P^2+P^4]=sin(2pi/7)+sin(4pi/7)+sin(8pi/7)
=-sin(pi/7)+sin(2pi/7)+sin(pi-3pi/7)=-sin(pi/7)+sin(2pi/7)+sin(3pi/7)=[root7]/2

 

[sonic1988]

i dont get y sin(8pi/7) = -sin(pi/7) instead of it equals sin(pi/7)

 

[ssglain]

i dont get y sin(8pi/7) = -sin(pi/7) instead of it equals sin(pi/7)

8pi/7 is in the third quadrant hence sin(8pi/7) has the negative value of the sine of related angle pi/7.

 

[sonic1988]

Oh yeh, my bad. I was confused with 7pi/7 which gives only half revolution. And yes, im doing 4 unit for hsc this year. I noe my question is stupid so yeh..