Proving Equality of Cubed Complex Numbers with Graphical Representation

[redount2k9]

We have a,b,c different complex numbers so

(a+b)^3 = (b+c)^3 = (c+a)^3

Show that a^3 = b^3 = c^3

From the first equality I reached a^3 - c^3 + 3b(a-c)(a+b+c) = 0 How a is different from c => a-c is different from 0
How do I show that a^3 - c^3 = 0?

 

[MrWarlock616]

If you can prove that 3b(a-c)(a+b+c)=0, you're done.
a, b, c are complex numbers.

 

[redount2k9]

But how to prove that?

 

[MrWarlock616]

Well, you can try assigning a value to the complex numbers and put them in the expression.

 

[micromass]

Well, you can try assigning a value to the complex numbers and put them in the expression.

Huh? How can that ever be a good proof??

 

[MrWarlock616]

lol i think it can't be..but you can do it if you take the time..
edit: expand the expression first.

 

[haruspex]

From the first equality I reached a^3 - c^3 + 3b(a-c)(a+b+c) = 0

That's a good start. (It will turn out that a+b+c=0.) If you take out the factor a-c, and also write down a corresponding equation with a/b/c rotated around, then add the two, what do you get?

 

[haruspex]

Doh! There's a much easier way.
We know (a+b) = (b+c)ω, where ω³=1. If a≠c then ω≠1. Writing the corresponding eqn for c+a versus b+c, and assuming b≠a and b≠c, we have (b+c) = (c+a)ω. (If (b+c) = (c+a)ω² then b=c.) It's not hard from there.

 

[MrWarlock616]

redount2k9, this formula can help you:

$x^3-y^3=(x-y)(x^2+xy+y^2)$

You will need to prove 3b(a-c)(a+b+c)=0 using this forumla, as I said earlier. Or you can put $a=x_1+iy_1, b=x_2+iy_2, c=x_3+iy_3$, but this will become lengthy.

 

[Curious3141]

MrWarlock616,

You will need to prove 3b(a-c)(a+b+c)=0 using this forumla, as I said earlier/QUOTE]

Hmmm. Using the above equation b=0 and a=c. Then a+b+c = a+c = 2a = 0 . Therefore a=b=c=0. Is that what you want to aver?

Ratch

xyz = 0 does NOT imply that x, y and z are all zero. It does, however imply that one or more of x, y and z are 0.

 

[MrWarlock616]

MrWarlock616,



xyz = 0 does NOT imply that x, y and z are all zero. It does, however imply that one or more of x, y and z are 0.

Yes, but doesn't it prove that $a-c=0$ and hence, $3b(a-c)(a+b+c)=0$?? I don't get what's the huge fuss is about.

 

[Joffan]

I don't know if the original poster is still with us...

We know a, b, and c are distinct.

Therefore (a+b), (b+c) and (c+a) are also distinct

Therefore , as per the question, they are the three distinct cube roots of some number z.

And therefore what is the sum ((a+b) + (b+c) + (c+a))?

 

[MrWarlock616]

2(a+b+c)

 

[Joffan]

More specifically, what is the sum of the three distinct cube roots of some complex number z?

 

[MrWarlock616]

That's 0 ..there you go. This can be locked. :P

 

[micromass]

Anyway. Use that if

(a+b)^3=(b+c)^3

then

a+b=e^{2ik\pi/3}(b+c)

for k\in \mathbb{Z}. And the same thing for the equality (a+b)^3=(a+c)^3.

Now solve the equations

\left\{\begin{array}{l} a+b=e^{2ik\pi/3}(b+c)\\ a+b=e^{2ik^\prime\pi/3}(a+c) \end{array}\right.

 

[micromass]

Thread is open again. Let's actually wait for the OP to reply before helping any further

 

[redount2k9]

What should I reply? All of you have different opinions... it's hard for me to understand something.

 

[MrWarlock616]

hahahaha use any of the methods..

 

[redount2k9]

Hahahaha why not to use the best method? (if I would know which it is)

 

[MrWarlock616]

I think what joffan said would be easiest! :)

 

[Curious3141]

OK, I'll post this hint. This method is essentially identical to what haruspex posted in post #8, but more explicit. Micromass reiterated the same solution in post #16, but it's easier to "grasp" with the omega symbol than with the exponential notation.

There are three cube roots of unity, $1, \omega, \omega^2$.

The original system can be reduced to:

$a + b = \omega(b + c)$

$b + c = \omega(c + a)$, which is equivalent to $a + b = \omega^2 (c + a)$

and these are solved simultaneously.

The system of equations has to be just so, because:

$a + b = (1)(b + c)$ would imply that a = c, violating uniqueness of a, b and c.

and

$a + b = \omega(c + a)$ would imply that $\omega (b + c) = \omega (c + a)$, giving a = b, again violating uniqueness.

So that's the only admissible system of equations, and they can easily be solved by substitution. Then using the well-worn algebraic properties of $\omega$ and $\omega^2$, e.g. $\omega^2 = -1 - \omega$ and $(1 + \omega)^3 = -1$, etc., the solutions can easily be manipulated to prove $a^3 = b^3 = c^3$.

Do you understand the outline of this solution? It's a lot less tedious than going the direct algebraic route.

 

[HallsofIvy]

Hahahaha why not to use the best method? (if I would know which it is)

The point is to do something yourself! If you do not know how to solve a problem try many different ways to see if one works or if the result of trying suggest something else. If you are given different ways to solve a problem and don't know which is "best", try one and see if it works. Even if it is not the best way, you will have solved the problem which is the entire point!

 

[Hurkyl]

The homogeneous equation (a+b)^3 = (a+c)^3 defines an algebraic curve of degree 3 in the projective plane (with indeterminates a, b, c).

So does the homogeneous equation (a+b)^3 = (b+c)^3.

Therefore, exactly one of the following two statements is true:

• The two curves have a common component, and thus there are infinitely many solutions
• The two curves intersect in 3*3 = 9 points in the complex projective plane, counting multiplicity.

We can check that the following gives 9 distinct projective solutions:
a = R \omega^m \qquad b = R \omega^n \qquad c = R
where \omega is a cube root of unity, R is a nonzero complex number, and m,n are integers. (projective solutions ignore scaling, so R=1 and R=2 would both give the same projective solution if m,n are kept the same)

If we can show that there aren't infinitely many solutions, then this, along with a=b=c=0, must be the complete solution set to the original equation.

 

[VantagePoint72]

Hurkyl, given the level at which the question is posed, do you really think even your first sentence is going to be intelligible to him or her? I don't see what the point is in providing a solution—even if it's correct—that is obviously enormously beyond the asker's academic level.

 

[ehild]

OK, I'll post this hint. This method is essentially identical to what haruspex posted in post #8, but more explicit. Micromass reiterated the same solution in post #16, but it's easier to "grasp" with the omega symbol than with the exponential notation.

There are three cube roots of unity, $1, \omega, \omega^2$.

The original system can be reduced to:

$a + b = \omega(b + c)$

$b + c = \omega(c + a)$, which is equivalent to $a + b = \omega^2 (c + a)$

and these are solved simultaneously.

The system of equations has to be just so, because:

$a + b = (1)(b + c)$ would imply that a = c, violating uniqueness of a, b and c.

and

$a + b = \omega(c + a)$ would imply that $\omega (b + c) = \omega (c + a)$, giving a = b, again violating uniqueness.

So that's the only admissible system of equations, and they can easily be solved by substitution.

See attachment. a+b, b+c and c+a are three cubic roots of the complex number z, all with magnitude r. They differ only in the angle of 120°.

Add two of them and subtract the third.

(a+b)+(b+c) = a+c+2b, (a+b)+(b+c)-(a+c)=2b.

You see from the picture that the yellow shaded triangle is equilateral, so a+c+2b is equal in magnitude with r. Also it makes the angle 180° with (a+c): So 2b=-2(a+c) You can apply the same procedure to get a and c in terms of the original complex numbers. I think, it is straightforward from here.

ehild

 

[Joffan]

You can actually pose this question with any exponent, with suitable sums.

For example:

Suppose a,b,c,d and e are distinct complex numbers, and that

(b+c+d+e)⁵=(a+c+d+e)⁵=(a+b+d+e)⁵=(a+b+c+e)⁵=(a+b+c+d)⁵

Show that a⁵=b⁵=c⁵=d⁵=e⁵

Don't do this by expanding equations ;-)

 

[Curious3141]

See attachment. a+b, b+c and c+a are three cubic roots of the complex number z, all with magnitude r. They differ only in the angle of 120°.

Add two of them and subtract the third.

(a+b)+(b+c) = a+c+2b, (a+b)+(b+c)-(a+c)=2b.

You see from the picture that the yellow shaded triangle is equilateral, so a+c+2b is equal in magnitude with r. Also it makes the angle 180° with (a+c): So 2b=-2(a+c) You can apply the same procedure to get a and c in terms of the original complex numbers. I think, it is straightforward from here.

ehild

Nice!

 

[redount2k9]

Thanks everyone but please understand that I'm only 14. I'll try to figure it out somehow.

 

[Joffan]

Thanks everyone but please understand that I'm only 14. I'll try to figure it out somehow.

Well, good luck. It will help if you are aware of the graphical representation of complex numbers and the idea that there are exactly three distinct cube roots of any complex number - and what they look like in the graphical version.