System of equations and recurrence relation I am trying to find the general solution for $N$ of the following system of equations
$$
\begin{cases}
    (x_n - x_{n-1})^2 + (y_n - y_{n-1})^2 = \left(\frac{\theta}{N}\right)^2 \\
    {x_n}^2 + {y_n}^2 = 1
\end{cases}
$$
with the initial values $x_0 = 1$ and $y_0 = 0$ and the following

*

*$\theta$ is a constant and $0 \leqslant \theta \leqslant 2$

*$N$ is a constant and we want to find the terms $(x_N, y_N)$
Using substitution with respect to $N$, we have
$$
\begin{align}
    x_0 = 1 \quad & ; \quad y_0 = 0 \\
    x_1 = -\frac{\theta^2 - 2N^2}{2N^2} \quad & ; \quad y_1 = -\frac{\theta \sqrt{4N^2 - \theta^2}}{2N^2} \\
    x_2 = \frac{\theta^4 - 4N^2\theta^2 + 2N^4}{2N^4} \quad & ; \quad y_2 = \frac{(\theta^3 - 2N^2\theta) \sqrt{4N^2 - \theta^2}}{2N^4} \\
    x_3 = -\frac{\theta^6 - 6N^2\theta^4 + 9N^4\theta^2 - 2N^6}{2N^6} \quad & ; \quad y_3 = -\frac{(\theta^5 - 4N^2\theta^3 + 3N^4\theta) \sqrt{4N^2 - \theta^2}}{2N^6}
\end{align}
$$
By using substitution, it becomes very difficult with $N \geqslant 2$.
 A: Using complex numbers as well, it is obvious from the second equation that the $n^{th}$ point can be written
$$z_n=e^{i\alpha_n}.$$
Then
$$|e^{i\alpha_n}-e^{i\alpha_{n-1}}|=|e^{i(\alpha_n-\alpha_{n-1})}-1|=|e^{i\alpha}-1|=\frac\theta N.$$
It is easy to show that this equation has at most two solutions in $e^{i\alpha}$, which are conjugate. If we always keep the same sign of $\alpha$, then by induction
$$z_n=e^{i\alpha_n}=e^{i\alpha_{n-1}}e^{i\alpha}=e^{in\alpha}.$$
A: The second equation expresses that the points $(x_n,y_n)$ remain on the unit circle, and the first, that the successive points form chords of constant length, subtending an angle $\alpha=2\arcsin\frac\theta{2N}$.
Hence $$(x_n,y_n)=(\cos n\alpha,\sin n\alpha).$$

By the way, the system has in fact $2^n$ distinct solutions, as from every intermediate point, the chord can be drawn in two directions.
A: Because it is very difficult to solve this relation, we can use complex numbers. Let
$$(x_n, y_n) = x_n + iy_n \quad \text{and} \quad z_n = x_n + iy_n$$
then we have
$$z_1 = -\frac{\theta^2 - 2N^2}{2N^2} + i\frac{\theta \sqrt{4N^2 - \theta^2}}{2N^2}$$
and the term $z_1$ is always given by the same expression with respect to $N$.
To compute the product of two complex numbers, we multiply the magnitudes and add the arguments. Since $\lvert z_1 \rvert = 1$, we have
$$\forall k \in \mathbb{N}^*, \lvert (z_1)^k \rvert = 1$$
and it rotates $z_1$ by $\arg(z_1)$ for $k$ times. Since the magnitude is always equal to 1, we stay in orbit around the unit circle.
Consider the following picture:

In particular, we notice that
$$\forall n,N \in \mathbb{N}^*, \arg(z_n) \leqslant \frac{\theta}{N} \quad , \quad n \leqslant N$$
because the length of the chord between two terms is given by
$$\frac{\theta}{N} = \lvert z_n - z_{n-1} \rvert = \sqrt{(x_n - x_{n-1})^2 + (y_n - y_{n-1})^2} \quad , \quad n \leqslant N$$
and we can reduce the length of each chord such that we obtain the term
$$z_N = \left(-\frac{\theta^2 - 2N^2}{2N^2} + i\frac{\theta \sqrt{4N^2 - \theta^2}}{2N^2}\right)^N$$
and by taking its limit, we eventually have
$$\cos(\theta) + i\sin(\theta) = \lim\limits_{N \to \infty} \left(-\frac{\theta^2 - 2N^2}{2N^2} + i\frac{\theta \sqrt{4N^2 - \theta^2}}{2N^2}\right)^N$$
as each chord becomes smaller and smaller and closer to the circle, we denote by $\theta$ the arc length measure around the unit circle in radians.
