Complex Number Vectors Questions?

Question

Solve the equations x^2 + x + 1 = 0 and x^2 - Sqrt3x + 1 = 0. Let A and B be the solutions to the first equation and C and D be the solutions to the second equation so that A and C lie above the real axis.

Find the angles:

i) AOB , ii) COD , iii) COA , iv) ACB

Answer 1

A=-0.5+i*(sqrt3)/2

B=-0.5-i*(sqrt3)/2

C=(sqrt3)/2+i/2

D=(sqrt3)/2-i/2

arg(A)=arctg(((sqrt3)/2)/(-0.5))=arctg... deg or 300 deg

We know A to be in the 2-nd quarter, so we stick with arg(A)=120 deg.

arg(B)=-arg(A)          (property of complex conjugates)

arg(C)=arctg(0.5/((sqrt3)/2))=arctg(1/... deg or 210 deg.

Knowing C to be in the 1-st quarter, we choose arg(C)=30 deg.

arg(D)=-arg(C)

To find the angles you are asked about in i, ii, iii, use the formula:

angle XOY=arg(Y)-arg(X)

Answer 2

We should first solve the equations :)

x^2 + x + 1 = 0

(x + 1/2)^27 = 1/4 - 1

x + 1/2 = +/- sqrt(-3) / 2 = +/- i sqrt 3 / 2

x = 1/2 +/- i sqrt(3) / 2

Ax = 1/2 + i sqrt(3) / 2

The angle of A is

A = tan-1 ( sqrt(3) / 1 ) = 60º

Bx = 1/2 - i sqrt(3) / 2

The angle is:

B = tan-1 (-sqrt(3)) = -60º

2---- x^2 - sqrt (3) x + 1 = 0 (Is it sqrt(3x) or sqrt(3) · x?? anyway u see how to do it)

(x - sqrt(3) / 2)^2 = -1 + 3/4 = -1/4

x = 1/2 +/- i /2

The solutions:

Cx = 1/2 + i /2

C = tan-1 (1) = 45º

Dy = 1/2 - i/2

D = tan-1 (-1) = -45º

Let's find the angles:

i) AOB

A has an angle of 60º and B has an angle of -60 therefore AOB = 60 -(- 60) = 120º

ii) COD

C = 45º

D = -45º

COD = 45-(-45) = 90º

iii) COA

C = -45º

A = 60º

COA = -45 - 60 = -105º

iv) ACB

This is tricky.

A, C and B has a real coordinate of 1/2:

.............. A

..............

.............. C <----- 180º

..............

..............

.............. B

ACB = 180º