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# complex numbers

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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

#1
+1

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
#2
+185
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could very well might be a chatgpt-generated solution so wanted to see other people's opinions on this

idontknowhowtodivide  Mar 17, 2023
#5
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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
#3
+397
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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
#4
+185
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yes i get x^2+x+1=0 but i don't know what to do next

idontknowhowtodivide  Mar 18, 2023