Continuation of Complex no. problems - finding an angle (1 Viewer)

[cssftw]

Hi, I'm having a problem with the absolute last question, Q1, (c) FIND angle(ACB)
Here's the question:

Q1. (a) If p is real . -2 (<) p (<) 2, show that the roots of the equation are non-real complex numbers with modulus 1.
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This was already answered in the other thread, I suppose the main thing from this question is that the MODULUS IS 1.

(b) Solve the equations x^2 + x + 1 = 0 and x^2 - x*(SQRT(3)) + 1 = 0. Plot on an Argand Diagram the points A and B representing the solutions of the first equation; and C and D representing the 2ND EQUATION. A and C LIE ABOVE THE REAL AXIS
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Ok, for EQUATION 1: x = [ -1 +/- i*sqrt(3)]/2

For EQUATION 2: x = [sqrt(3) +/- i]/2

Plotting them on the Argand Diagram is simple enuff...


(c) Find the angles: AOB, COD, COA, ACB

angle(AOB) = arg(B) - arg(A)]
= pi + pi/3
= 2pi/3

angle(COD) = arg(C) - arg (D)
= pi/6 + pi/6
=pi/3

angle(COA) = arg(A) - arg(C)
= (pi - pi/3) - pi/6
= pi/2

Now the problem I've been having:

Find angle(ACB)
I really got no idea what to do, so the help of you guys would really be welcome.

Sorry, guys, you're gonna have to draw this question out :L

thanks guys

 

[ribosome911]

This is what I did:

EQUATION 1: x^2 + x +1 = 0 , since all coefficients are real, let 2 complex roots be rcis(theta), rcis(- theta).
modulus of root =1 , therefore r=1.
From relations of sum of roots and coefficients, cis(theta) + cis ( - theta) = -1.
cos(theta) +isin(theta) +cos(theta) - isin(theta) = -1.
2cos(theta) = -1.
cos(theta) = -1/2.
theta= cos^(-1) (-1/2).

If you do the same for x^2 - x* (SQRT 3) + 1, you shoud be able to get cos(theta) = (SQRT 3)/2, hence theta = pi/6.