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I start by drawing a line of 1 unit on the $x$ axis. I turn left (from the perspective of an ant on the line) by an angle of $\alpha$ and I draw a second segment of length $u$ from my endpoint of the first segment. I then turn another angle $\alpha$ and then draw a $3^{rd}$ segment of length 1, etc. alternating between segments of length $1,u,1,u,...$ and rotating by a fixed angle each time. I am interested in when does this process returns to the origin and how long does it take for it to return when it does.

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closed as off-topic by Matthew Conroy, Claude Leibovici, JonMark Perry, user91500, Arnaud D. Sep 19 '17 at 20:13

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We work in the complex plane. Let $z=e^{i\alpha}$. Then the path is: $$1+uz+z^2+uz^3+z^4+uz^5+\cdots$$ If $1+uz=0$, meaning $u=1$ and $\alpha=\pi$, then we return to $0$ after $2$ steps. Otherwise, assume that $1+uz\neq0$.

Suppose the path returns to $0$ after $2N$ steps for $N>1$. Then we can solve for this: $$0=(1+uz)\sum_{k=0}^{N-1}z^{2k}$$ Divide out $(1+uz)$: $$0=\sum_{k=0}^{N-1}z^{2k}$$ This is true if and only if $z^{2N}=1$ and $z^2\neq1$. In other words, $\alpha$ is a multiple of $\pi/N$, but not $0$ or $\pi$. Note that $u$ is irrelevant!

Tackling an odd number is harder. Suppose the path returns to $0$ after $2N+1$ steps. Then: $$0=z^{2N+1}+(1+uz)\sum_{k=0}^{N-1}z^{2k}=z^{2N+1}+(1+uz)\frac{1-z^{2N}}{1-z^2}$$ $$0=z^{2N+1}-z^{2N+3}+1+uz-z^{2N}-uz^{2N+1}$$ That's a really messy equation, and I'm not sure if there's a simple class of solutions.

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The named system of equations solving for $n=2k,\alpha,u$ should be: $$ u \sum_{i=0}^k \cos((2k+1)\alpha)+\sum_{i=0}^k \cos(2k\alpha)=0\\ u \sum_{i=0}^k \sin((2k+1)\alpha)+\sum_{i=0}^k \sin(2k\alpha)=0 $$ Hence the solution for a given $k$, if exists, is given by: $$ u = -\frac{\sum_{i=0}^k \cos(2k\alpha)}{\sum_{i=0}^k \cos((2k+1)\alpha)}\\ \frac{\sum_{i=0}^k \cos(2k\alpha)}{\sum_{i=0}^k \cos((2k+1)\alpha)} = \frac{\sum_{i=0}^k \sin(2k\alpha)}{\sum_{i=0}^k \sin((2k+1)\alpha)} $$ Which as you see, would have a closed form for some circular symmetry values for $\alpha$....

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The loci are polygons containing start point on the circle as we know the sum of external angles $=360^0$ for a full closed regular polygon. In general it would be infinite sided irregular polygon. The "polygon" can be drawn even on BASIC.

Diameter $D$ of approximate circum-circle so formed $ =\dfrac{1+u}{2 \alpha} $, Full Arc length $=\pi D$ to return to the start point as shown.

It can be seen that when $\alpha$ is small the locus tends to a circle.

When $n$ is integer, rotation angle at each vertex of polygon $\alpha =2\pi/(2n) $ when repeated forms a regular polygon.

Circles Tgt to X-axis

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