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So, I'm trying to figure out those coordinates, for the last 2 hours.

I have a point $x$ and $y$ on the cartesian plane, and i want the $N$ other points that are the "projection" of the first point every $2\pi / N$.

Example.

Let's say i have $[x,y] = [1,0]$ and $N = 5$, the 5 points that i'm interested in are the vertices of the pentagon with "the tip" on $[1,0]$, so I suppose in this case it should be:

$$\text{points} = [\cos( (2 p\pi / N)p k), \sin( (2 \pi / N) k)]$$

but I need of a "general" formula.

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  • $\begingroup$ These points like on an argand diagram with position$$(x+iy)e^{i 2\pi k/n}\qquad k=0,1,\dots,n-1$$Calculate the real and imaginary parts of this number to get the required position. $\endgroup$ Aug 21, 2020 at 21:17
  • $\begingroup$ @PeterForeman thank you... are there more "real" solutions? $\endgroup$ Aug 21, 2020 at 21:18
  • $\begingroup$ I suppose you could also just use a real rotation matrix with the given angle on the position vector $(x,y)^T$. $\endgroup$ Aug 21, 2020 at 21:19

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\begin{align} x_k &= \cos\left(\frac{2\pi}{n} k\right) x_0 - \sin\left(\frac{2\pi}{n} k\right) y_0 \\ y_k &= \sin\left(\frac{2\pi}{n} k\right) x_0 + \cos\left(\frac{2\pi}{n} k\right)y_0 \end{align} where $(x_0, y_0)$ is your starting point, I've used $n$ instead of $N$, and $(x_k, y_k)$ are the coordinates of the $k-$th point, so you'll need to plug in $k = 1, 2, 3, \ldots, n-1$.

This won't work if $(x_0, y_0) = (0, 0)$ ... or it will work, but all $n$ points that you get will be at the origin, so it won't be interesting.

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  • $\begingroup$ maaaaan is you mathematicians are ammmmmmaizing, thank you so much $\endgroup$ Aug 21, 2020 at 21:30
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    $\begingroup$ Glad to be of some help. $\endgroup$ Aug 21, 2020 at 21:32

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