A Quick Solution Using Complex Numbers

[@nirajmehta4264]

so the total product of all diagonals is n^2, accounting for each vertex. i imagine if someone told me that fact, i'd not be sure how to prove it. fascinating, another great video Dr. Barker. i always look forward to your videos!

[@adrienanderson7439]

correct me if I'm wrong, It seems like n^n would be the result of multiplying by each diagonal and edge exactly two times. This suggests that if we multiply by each diagonal between vertexes including the edges once we would get sqrt(n^n).

[@DrBarker]

Thank you!

[@DrBarker]

@@adrienanderson7439 I think this should be n^(n/2), since there are n vertices, and the product is n from each vertex, but each edge/diagonal is counted twice.

[@Sasseater]

​@@DrBarkerspot on

[@adrienanderson7439]

@@DrBarker I could have worded my original comment better but I was thinking about the product of all of the lengths of all of the diagonals and edges. I think that you are talking about the number of diagonals and edges. By the way I really like this video for all of the complex number techniques it shows.

[@RGP_Maths]

When I encountered this problem just over 9 years ago, I solved it in a very similar manner, which I think is even simpler. We want the product of |z-1|, taking the product over all z for which z^n=1, except z=1. As you say in the video, this is the same as |product (z-1)|. Let w = z-1, so we need the product of all w for which (w+1)^n=1, except w=0. Expanding the binomial and then factorising we have: w [w^(n-1) + n w^(n-2) + ... + n] = 0. The non-zero w are the roots of the polynomial in the square bracket, and by Vieta's formulae their product is ±n. But recall that we need the modulus of this product, which is n.

[@armanavagyan1876]

Thanks PROF pretty interesting and useful)

[@jacemandt]

Since the product has to be n, and since there are symmetrical chords of the circle, the lengths (if they are algebraic) are constrained to contain √k, where k is a factor of n. This is why lengths in regular pentagons contain √5, lengths in squares and regular octagons contain √2, and lengths in equilateral triangles and regular hexagons contain √3.

[@DavidSartor0]

This is a great result. Thank you for the proof; I hadn't known one. Just yesterday, I gave this problem to my Linear Algebra teacher.

[@themaverick1891]

Thank you so much Dr. Barker. I get to learn a lot for you. I admire your knowledge and passion for teaching. Bless you. 😄

[@abdallahmo8385]

A great example of generating functions!

[@synaestheziac]

This is stunningly beautiful, and I’m sort of shocked that I’ve never seen it before! If you don’t mind me saying, I think the question should be reworded a bit. First, it should say “vertex of”instead of “point on”. Second, it should ask for the product of all diagonals *and edges*, not just all diagonals, right? Anyway, thanks for another great video!

[@hubertorhant8884]

Brilliant ✨

[@Unchained_Alice]

I knew the roots of unity lie on a unit circle and trace out an n-gon but never thought of doing this!

[@Hipeter1987]

So an equilateral triangle inscribed in the unit circle has side lengths sqrt(3).

[@greenpewdiepie4207]

Interestingly, for any n-gon inscribed in a circle of radius r, the product of all diagonals from an arbitrary point to all other points on that n-gon is nr^(n-1). This is curiously equivalent to the derivative of r^n with respect to r. Is there any connection?

[@firstnamelastname307]

surprising result

[@guruone]

Nice

[@mcwulf25]

Clever!

[@sunsetclub4132]

Why can we do z^8 = 1? Is it something to do with the area of the n-gon?

[@shreyas713]

In the complex plane, graphing the solutions to z^8 = 1 would display a regular octagon, which is the polygon we would be working with. To generalize this, graphing the solutions to z^n = 1 in the complex plane (and connecting the points adjacent) gives you a regular n-gon.

[@sunsetclub4132]

@@shreyas713 Did not know that, thank you very much!

[@Mephisto707]

5:24 I couldn’t grasp why z=1 suddenly became valid.

[@iabervon]

When we made (z^n-1)/(z-1), this rational function is equal to the original polynomial except that it has a hole at 1 where the original was continuous. Then we found a second polynomial that's equal to the rational function except that it's continuous at 1. We don't know directly that the value at 1 is the same between the two polynomials, but since they're both continuous and are the same everywhere else, they must have the same limit at 1 and therefore the same value. If you poke a hole in a continuous function and that fill it in to get a continuous function, you must get the same function you started with. (And all polynomials are continuous.)