Complex number question stuck (1 Viewer)

[LaCe]

Q4. (13 marks – 2, 3, 4, 4)

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

(i) Prove that p^6 + p^5 + ... + p + 1= 0

(ii) The second root of the quadratic equation is (beta). Express (beta) in terms of positive powers of p . Justify your answer.

(iii) Find the values of the coefficients of a and b.

I have done i) and ii) but iii) bothers me

I have 2 equations to find (alpha) and (beta): product of roots=b, sum of roots=a

to find a: u get a= p^6+p^5=...+p
therefore a=-1 from i)

how do u get b?? ive tried subbing in cis values but u end up with = 2cos(2pi/7) + 2cos(4pi/7) + 2cos(6pi/7)

Cheers

 

[KFunk]

for iii) if a and b are real then the conjugate of p is also a root. (i'm assuming that you intended a plus sign, not an equals)

Then you could use (x - p)(x - p_conjugate)=(x² + 2Re(p)x + |p|²)

which yields:

a = 2.[cos(2π/7)]

b= [cos(2π/7)]² + [sin(2π/7)]² = 1 ... (strangely enough)

 

[LaCe]

Actually, that's not the answer needed. I dont think u can do it by that method

when u find the two roots (alpha) and (beta) in terms of p

u use (alpha)+(beta)=a
(alpha)*(beta)=b

then as the coefficients of the quadratic are real, (alpha conjugate) is a root

a=-1 by i)
b is suppose to equal 2

?

 

[maths > english]

The question from taken from question 8 of the 1999 hsc. You got the definition of alpha wrong. Its meant to be p+p^2+p^4. Solution attached

 

[LaCe]

oh thanks, i forgot about roots go + - +

thanks