Quick complex numbers problem (1 Viewer)

[petergriffin198]

Help with part (iv) pls.
Btw is this under the topic complex numbers or polynomials?

 

[snail489]

View attachment 32436

Help with part (iv) pls.
Btw is this under the topic complex numbers or polynomials?

In part (iii) you find a=1 and b=2 (or at least that's what I got), so then you use the quadratic x^2+x+2=0 and solve it for the roots alpha and beta. Then you equate the imaginary parts so you can get that sin expression thing.

 

[physchemnotes]

you substitute the values of a and b back to the original equation, x^2 - x + 2 =0. Then you use the quadratic formula to find Im(x) = +-root(7)/2.

From here we know that beta (q^3 + q^5 + q^6) must be the case with the negative imaginary values, by the nature of the way we go around the unit circle. So, taking the imaginary values of each side we get sin(6pi/7) + sin(10pi/7) + sin(12pi/7) = -root(7)/2. Then its a simple matter to equate different sine values and get the correct answer.

Sorry Im pretty bad at explaining hope that helped