Solve Complex Equations | Helpful Tips for Troublesome Equations

[Ed Quanta]

Can someone help me solve this equation which for some reason is giving me trouble?

z^2+z+1=0 where z is a complex number, or if it makes it easier we can write

(x,y)(x,y) + (x,y) + (1,0)= (0,0)

 

[fourier jr]

just use the quadratic formula. You'll get 2 roots because z can be complex.

 

[Ed Quanta]

I was trying to do it by separating equations. Anyway, thank you, do you know where I can find an online proof of the quadratic formula?

 

[Gokul43201]

Proof lies in completing the square.

$$ax^2 + bx + c = 0 = {x^2} + \frac {b} {a} x + \frac {c} {a}$$
$$(x + \frac {b} {2a})^2 - \frac {b^2} {4a^2} +\frac {c} {a} =0$$

Take constant terms to other side, find the square root and subtract b/2a to get the quadratic formula.

 

[kuenmao]

I'll just type that for you: (the "^" means "to the power")

a x^2 + b x + c = 0
x^2 + (b x)/a + c/a = 0 (divide both sides by a, since a is not zero)
x^2 + (b x)/a = -c/a
x^2 + (b x)/a + (b/2a)^2 = -c/a + (b^2)/(4 a^2) (Add (b/2a)^2 to both sides)
By the identity a^2 + 2ab + b^2 = (a+b)^2, we have
(x + b/2a)^2 = -c/a +(b^2)/(4 a^2)
(x + b/2a)^2 = (b^2-4ac)/(4 a^2)
x + b/2a = sqrt.[b^2-4ac] / 2a or -sqrt.[b^2-4ac] / 2a
x = (-b + sqrt.[b^2-4ac]) / 2a or (-b - sqrt.[b^2-4ac]) / 2a

And so you have the quadratic formula. Hope that helps!

 

[kuenmao]

whoops, I am new here and just realized that you could create the equations yet...sorry about making that long chunk in the previous post!

 

[Gokul43201]

whoops, I am new here and just realized that you could create the equations yet...sorry about making that long chunk in the previous post!

Welcome to PF kuenmao...check out the LaTex post under General Physics.

 

[robert Ihnot]

the roots of unity

Another way to look at that is (z-1)(z^2+z+1) = z^3 -1. Thus we are talking about the three roots of 1. Those roots are cos(k(120)) + isin(k(120)), for k=1,2,3. This actually is a better way to do it from the standpoint of insight into the roots of unity. However it is not a general method for solving the quadratic.