Problem source

A path is made up in the Argand diagram of a series of straight line segments $P_{1}P_{2},$ $P_{2}P_{3},$ $P_{3}P_{4},\ldots$ such that each segment is $d$ times as long as the previous one, $(d\neq1)$, and the angle between one segment and the next is always $\theta$ (where the segments are directed from $P_{j}$ towards $P_{j+1}$, and all angles are measured in the anticlockwise direction). If $P_{j}$
represents the complex number $z_{j},$ express 
\[
\frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}}
\]
as a complex number (for each $n\geqslant2$), briefly justifying your answer. 
If $z_{1}=0$ and $z_{2}=1$, obtain an expression for $z_{n+1}$ when $n\geqslant2$. By considering its imaginary part, or otherwise, show that if $\theta=\frac{1}{3}\pi$ and $d=2$, then the path crosses the real axis infinitely often.

Solution source

\begin{align*}
&& | \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} | &= d \\
&& \arg \left (  \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} \right) &= \arg (z_{n+1}-z_{n}) - \arg(z_{n}-z_{n-1}) \\
&&&= \theta \\
\Rightarrow && \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} &= d e^{i \theta}
\end{align*}

\begin{align*}
&& z_1 &= 0 \\
&& z_2 &= 1 \\
&& \frac{z_3-z_2}{z_2-z_1} &= de^{i \theta} \\
\Rightarrow && z_3 &= de^{i \theta} + 1 \\
&& \frac{z_4-z_3}{z_3-z_2} &= de^{i \theta} \\
\Rightarrow && z_4 &= (d e^{i \theta})^2 + d e^{i \theta}  + 1\\
\Rightarrow && z_{n+1} &= \frac{(de^{i \theta})^{n}-1}{de^{i \theta}-1}
\end{align*}

If $d = 2, \theta = \tfrac13 \pi$, then, $2e^{i \tfrac13 \pi} = 1 + \sqrt{3}i$

\begin{align*}
\textrm{Im}(z_{n+1})) &= \textrm{Im} \left ( \frac{(2e^{i \tfrac13 \pi})^{n}-1}{2e^{i \tfrac13 \pi}-1}\right) \\
&= \textrm{Im} \left ( \frac{(2e^{i \tfrac13 \pi})^{n}-1}{\sqrt{3}i}\right) \\
&= -\frac{1}{\sqrt{3}}\textrm{Re} \left (2^n e^{i \frac{n}{3} \pi} \right) + \frac{1}{\sqrt{3}}
\end{align*}

Which clearly changes sign infinitely many times, ie crosses the origin infinitely many times.