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complex numbers question

Write down the necessary and sufficient conditions that the distinct complex numbers a,b,c represent the vertices of a equilateral triangle taken in clockwise order.

So I'm not really too sure how to go about this. I drew a sketch to get an idea, and I know that if it's an equilateral triangle then the magnitude of the lengths are the same, and we will have to rotate by -pi/3 to go clockwise...
Original post by maths learner
Write down the necessary and sufficient conditions that the distinct complex numbers a,b,c represent the vertices of a equilateral triangle taken in clockwise order.

So I'm not really too sure how to go about this. I drew a sketch to get an idea, and I know that if it's an equilateral triangle then the magnitude of the lengths are the same, and we will have to rotate by -pi/3 to go clockwise...


look at a basic equilateral triangle (say one with centre the origin and one vertex at 1) and see if you transform it into a general equilateral triangle


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Reply 2
Original post by maths learner
Write down the necessary and sufficient conditions that the distinct complex numbers a,b,c represent the vertices of a equilateral triangle taken in clockwise order.

So I'm not really too sure how to go about this. I drew a sketch to get an idea, and I know that if it's an equilateral triangle then the magnitude of the lengths are the same, and we will have to rotate by -pi/3 to go clockwise...


THe complex a,b and c will be the solutions of

z3=1\displaystyle z^3=1 equation

In this case the midpoint of the triangle will be in the origin and the magnitude of the length is 3\sqrt{3}

For midpoint in z0z_0 and length of k you shoul to solve

(zz0)3=k3\displaystyle \left (z-z_0\right )^3=\frac{k}{\sqrt{3}}
(edited 9 years ago)
demorphes theorem. Modulus of 1 hence rotion o x radians! e^ix=cos(x) +isin(x)


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Original post by physicsmaths
demorphes theorem.


His name has certainly morphed.
Original post by Mr M
His name has certainly morphed.


lol i was in a hurry so !
de moivre's * lol happy haha


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Hmm... 5 responses and I can't really say I think any of them help very much (sorry guys).

I think the key observations here are: if you consider two complex numbers p, q to represent vectors, then:

|p/q| is the ratio in their lengths. So if you want them to be the same length, then we need |p/q| to equal ...?

arg(p/q) is the angle between the vectors (measuring clockwise from p to q). So if you want them to be two sides of a triangle, we need arg(p/q) to equal ...?

Finally, observe that p/q=p/qeiarg([/q)p / q = |p/q| e^{i\arg([/q)}, so if you know |p/q| and arg(p/q), then you know what p/q has to be....

[You might also want to observe that if a,b,c are the 3 points on your triangle, then the vectors representing the sides are b-a, c-b, a-c].
(edited 9 years ago)
Original post by DFranklin
Hmm... 5 responses and I can't really say I think any of them help very much (sorry guys).

I think the key observations here are: if you consider two complex numbers p, q to represent vectors, then:

|p/q| is the ratio in their lengths. So if you want them to be the same length, then we need |p/q| to equal ...?

arg(p/q) is the angle between the vectors (measuring clockwise from p to q). So if you want them to be two sides of a triangle, we need arg(p/q) to equal ...?

Finally, observe that p/q=p/qeiarg([/q)p / q = |p/q| e^{i\arg([/q)}, so if you know |p/q| and arg(p/q), then you know what p/q has to be....

[You might also want to observe that if a,b,c are the 3 points on your triangle, then the vectors representing the sides are b-a, c-b, a-c].


Yeah i was trying it the last way you mentioned. Would it just be ba=caeiπ3 |b-a|=|c-a|e^{-\frac{i\pi}{3}}
Nope. |b-a| and |c-a| are both real numbers, while e^(-i pi / 3) is not.
Original post by DFranklin
Nope. |b-a| and |c-a| are both real numbers, while e^(-i pi / 3) is not.


But doesn't e^(-i pi / 3) represent a clockwise rotation of 60 degrees... Which is needed to get to each vertex?
Original post by maths learner
But doesn't e^(-i pi / 3) represent a clockwise rotation of 60 degrees... Which is needed to get to each vertex?
That might make sense if the other terms were directions (vectors), but they are not, they are distances. You are basically saying "rotate one distance by 60 degrees to get the other distance". Distances don't work like that!
Original post by DFranklin
That might make sense if the other terms were directions (vectors), but they are not, they are distances. You are basically saying "rotate one distance by 60 degrees to get the other distance". Distances don't work like that!


Ah okay. So if i removed the mod signs and had just (b-a)=(c-a)e^(-i pi / 3) it would work?
Original post by maths learner
Ah okay. So if i removed the mod signs and had just (b-a)=(c-a)e^(-i pi / 3) it would work?
In principle. I would check it with a canonical equilaterial triangle (e.g. origin, 1, (1+i \sqrt{3})/2 ) to make sure there are no sign errors.

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