Let a,b, and c be distinct integers and let x be a complex number such that x^3=1 and x≠1. Solve for the minimum value of the magnitude of a+bx+cx^2

 Mar 17, 2023

To solve for the minimum value of the magnitude of a+bx+cx^2 where a,b, and c are distinct integers and x^3=1 and x≠1, we can use the fact that x^3=1 to write x^3-1=0, which can be factored as (x-1)(x^2+x+1)=0. Since x is not equal to 1, then the equation x^2+x+1=0 must hold true.

Next, we can use the fact that x is a complex number to write x in terms of its real and imaginary parts, as x=r(cosθ+isinθ), where r is the magnitude of x and θ is the angle that x makes with the positive real axis. Substituting x into the equation x^2+x+1=0 and using Euler's formula, we can write the equation as:

r^2(cos2θ + cosθ) + r(sin2θ + sinθ) + 1 = 0

This is a quadratic equation in r, so we can solve for r using the quadratic formula:

r = [-b ± sqrt(b^2 - 4ac)] / 2a

where a = cos2θ + cosθ, b = sin2θ + sinθ, and c = 1. Note that a,b, and c are all real numbers.

Since we want to find the minimum value of the magnitude of a+bx+cx^2, we can write this expression in terms of r and θ and then minimize it with respect to θ. Using the formula for x and simplifying the expression, we get:

|a+bx+cx^2| = |a + brcosθ + crcos2θ + brsinθ + crsin2θ|

We can then use the fact that sin2θ = 2sinθcosθ and cos2θ = cos^2θ - sin^2θ to rewrite this expression as:

|a+bx+cx^2| = |(a + bcosθ) + (bsinθ + ccosθ)r + csin^2θ|

This is minimized when θ=pi/4, to give a minimum value of 2*sqrt(2).

 Mar 17, 2023

could very well might be a chatgpt-generated solution so wanted to see other people's opinions on this

idontknowhowtodivide  Mar 17, 2023

The answer is plain wrong. I have no idea what kind of sorcery they used to get \((r(\cos\theta+i\sin\theta))^2=r^2(\cos2\theta+\cos\theta)\), but it definitely was not "Euler's formula". Also, we are minimizing the value by adjusting a, b, and c, not \(\theta\)  as that is already determined, in fact it equals 120 or 240. Probably chatgpt.

Guest Mar 18, 2023

The roots of the equation are x^3 = 1 are 1, -(1/2) - I root(3)/2  and -(1/2) + I root(3)/2.

x =1 is excluded so x is one of the other two.

It doesn't matter which, since one is the square of the other.

What happens after that ....... .

 Mar 17, 2023

yes i get x^2+x+1=0 but i don't know what to do next

idontknowhowtodivide  Mar 18, 2023

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