a nice geometry problem in the complex plane.

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[goodplacetostop]

0:17 Since some people asked, here's the original problem: For some complex number ω with |ω| = 2016, there is some λ > 1 such that ω, ω², λω form an equilateral triangle in the complex plane. Then, λ can be written in the form (a+√b)/c, with a, b, c positive integers. Compute √(a + b + c).
9:21 My pleasure 👍

[pizzamidhead]

a nice thing is that if |ω| is √2 then λ is the golden ratio φ.

[jimschneider]

@9:20 - the fact that lambda is a real number greater than 1 means that the expression you derived for lambda also puts conditions on the modulus of omega. In particular, the expression under the radical must be greater than one, which means that the modulus of omega must be greater than one.

[genericcandy]

i think that your answer is too complicated, once you have the triangle with vertices 1, λ and ω, since all sides have the same length by definition, |ω - 1| = λ - 1, so λ = 1 + |ω - 1|. While the 2 formulas for λ don't seem to match at first, since arg(ω - 1) = ± π/3 (again, by looking at the triangle), i think they must match

[bmenrigh]

I especially liked this video because in part because of the simplicity of it despite being in the complex plane. I’d love to see more videos that use the complex numbers and coordinate geometry seems like a nice application.

[fboulben]

3 complex numbers a, b, c form an equilateral triangle iff a +bj +cj^2=0 where j^3=1 (j different from 1) . That gives you lambda right away

[davidgillies]

Note you can't just pick an arbitrary ω, plug it into λ and get an equilateral triangle.

[larspos]

An other formula that works: λ = 1+2/root(3)*Im(ω). Beacause not all ω wil give a value for λ, different formulae can, and hence do exist.

[xCorvusx]

Are you sure ω can lie in any quadrant?
It seems to me that, if its argument is greater than or equal to π/2 (at least), its square is too far around the origin to form an equilateral triangle the stated way.
Edit: In fact, the sketch at 4:22 seems to imply that θ must be less than π/3, since 1 and λ lie on the real axis and ω is supposed to be non-zero.

[unflexian]

the thing is an equilateral triangle cannot be created for all ω, only for ω of the form t+(√3t-√3)i. your expression for λ is correct but only defined on those values of ω.

[AnCoSt]

Carnegie Melon and her sister schools, Carnegie Cantaloupe and Carnegie Honeydew (kidding around, great video)

[udic]

If you are using "well known facts" about complex multiplication i think you can use a well known fact about 30°:60°:90° triangle with sides 1:sqrt(3):2 instead of the use of trigonometric functions.

[akaRicoSanchez]

Reached the same result through another way : I expressed the side of the equilateral triangle using the cosine theorem in the (origin, omega, omega squared) triangle as I know two sides of the triangle and know the angle at omega must be 120°.

[Karatemaci]

Beautiful! And because j has two complex values, all the time we have two solutions as the geometry clearly shows.

[FreshBeatles]

im so happy whenever we get a geometry problem

[nerdiconium]

I used the fact that ohm^2-ohm and lambda*ohm -ohm are 60 degree rotations of each other, used two sixth roots of unity, solved from there, and solved for the conditions when lambda was real.

[Hyakurin_]

Equating the three sides, i got lambda(w)=2Re(w)-1, so Re(w)>1 under the condition Im(w)^2=4Re(w)^2+2Re(w)-1

[tomatrix]

In your fact, it would be clearer to say the distance is multiplied by |z|, otherwise ‘changes by’ would suggest addition which isn’t true necessarily.

[anon]

I'm gonna call omega, w.
And lambda, k.
(edit: I know the problem wasn't to find w... But I was more interested in finding the triangles themselves than lambda)

k = 1 + sqrt ( ww* + 1 - 2a ) [where a is the real portion of w]
This value for lambda equates |ww-w| = |kw-w|, but not the third side.
If you plug this value of k (lambda) into |ww-kw|, you find that the imaginary portion is determined by the real portion.
This implies, the allowed values for w are on a line?
I find this line to be:
yy = xx - 1 [I'm calling them x, y but they're really the real/imaginary axes]
Using this to update k ...
k = 1 + sqrt (2aa-2a)

I find only two possible triangles:
The null-triangle: (all side-lengths are zero)
@ w = 1 + 0i
And the special(?) triangle: (side length = sqrt(28))
@ w = 2 + root3 i

Are there others?
(I'm seriously not confident in this answer)

[blackkk]

This relation holds only if |ω| > 1.