A Neat Application of Complex Numbers

We find the total length of all edges and diagonals of a regular n-sided polygon, inscribed in a unit circle. This involves a very nice use of complex numbers.
00:00 Intro
00:19 Edge lengths
03:13 Diagonal lengths
06:08 Diameter
07:25 Reflex angles
09:13 Sum of lengths
11:52 Using complex numbers
16:23 Making the denominator real
20:09 Taking the imaginary part

[@pableraspfgpfg468]

Such a beautiful problem for some minor olympiads. Easy to visualize, and not too complex to compute. Brilliant

[@williamperez-hernandez3968]

The final answer can be simplified further. Since sin(pi/n) = 2 sin (pi/2n) cos (pi/2n), and 1 - cos (pi/n) = 2 [sin(pi/2n)]^2, we get [sin (pi/n) ] / [ 1 - cos (pi/n)] = [cos (pi/2n)] / [sin (pi/2n)] = cot (pi/2n). Total = n cot (pi/2n).

[@lenskihe]

Nobody likes cot

[@worldnotworld]

Aww, that's mean@@lenskihe

[@lenskihe]

@@worldnotworld I just think that cot is generally unnecessary and doesn't help any kind of understanding. It is a neat simplification though, just write n / tan(π/2n) next time ;)

[@DrBarker]

Very neat!

[@pietergeerkens6324]

I went a different route and recognized the final formula as n * sin (pi/n) / versin (pi/n)

[@Adventurin_hobbit]

That was an elegant use of complex numbers.

[@kappascopezz5122]

The way I like to calculate the opposite length of an isosceles triangle is that I draw a line from the corner with angle θ to the center of the opposite side L. Due to symmetry, we know that this line is at a right angle to L, so it forms two mirrored right angled triangles together with θ/2 and L/2, since the new line cuts θ and L in half. And this right angled triangle lets you directly apply the sin=opposite/hypotenuse definition so you get L/2 = r sin(θ/2), which is the result that you calculated a few times in the video using trig identities. Before taking the geometric sum, you also could have added (e^(iπ/n))^0 to the start because that doesn't change the imaginary component, which makes the common ratio just 1 instead of e^(iπ/n), so you probably need to simplify less.

[@DrBarker]

Nice - yes, splitting into two right-angled triangles could have saved us some effort with the algebra using trig identities.

[@elephantdinosaur2284]

Nice video. Based on the title I thought you were going to use the roots of unity as the vertices of the polygon and work out the formula from 2 x Total = sum_{j,k} |z^j - z^k|. It was a great use of trig formulas and geometric reasoning without just brute forcing the algebra.

[@RGP_Maths]

So did I. That would certainly be the efficient way to prove that the product of the lengths of all those edges and diagonals is n^(n/2).

[@d.h.y]

The magic of complex numbers🤩🤩

[@jacemandt]

Nice calculations! The problem reminds me of one of my favorite theorems ever-maybe you could make this video too? Let V_n = the set of vertices of a regular n-gon inscribed in a unit circle. Choose any one point from that set, and draw the (n-1) chords connecting it to all the other points of the set. Then the product of the lengths of those chords is always (almost incomprehensibly) n.

[@DrBarker]

I actually came across this result when I was looking into making this video and thought it was really cool - it's a great suggestion! It would be nice to cover this result at some point.

[@DavidSartor0]

Where is there an explanation of this?

[@JohannNeumann-ic3gy]

Just found the channel through Dr. Trevor Bazett. Glad I did.

[@Nim2]

Thats really cool. Thanks

[@dneary]

I find it easier when calculating the length L to take the perpendicular bisector of the angle - you get L/2 = sin(theta/2) - and in the case where the angle is 2k pi/n this clearly gives L = 2 sin(k pi/n) without using angle identities.

[@General12th]

Hi Dr. Barker! So good!

[@NoNameAtAll2]

can't we just split isosceles triangles in half and apply sinus directly?

[@DrBarker]

Yes, this is a good idea! It should mean we don't need to do as much simplifying.

[@GhostyOcean]

Using the law of sines is the long way. Use the amplitude of the triangles to see L/2 = cos(pi/2 - pi/n) so L = 2sin(pi/n).

[@meraldlag4336]

Great problem and explanation

[@worldnotworld]

Wow. I had no idea complex numbers would appear in this way; I'd at first imagined the solution would use vectors represented as complex numbers corresponding to the relevant lengths in the complex plane, based on a circle |z|=1, which would have been perfectly Barkerian (and maybe there is such an approach?). The divergence into Im{...) seemed tedious at first, but the result is incredibly satisfying!

[@NotthatRossKemp]

Great little problem!

[@firstnamelastname307]

Amazing solution to the problem Dr! Did you come up with the problem too? Or is it a known one?

[@DrBarker]

Thank you! This is based on things I've seen before. I've seen various problems before where we can evaluate sums of trig functions using geometric series and real/imaginary parts of complex numbers. It was particularly satisfying to be able to combine this with the geometric setup though, rather than it just being a sum of trig functions without a geometric meaning.