Distance from center of a spiral Consider the following curve:


*

*We start at the center and take a step of 1 unit length

*Then turn to the left by $\pi/3$ and take a step of $9/10$ units in length

*Then turn to the left by $\pi/3$ and take a step of $81/100$ units in
length

*Then turn to the left by $\pi/3$ and take a step of $729/1000$ units in length 

*Repeat the left turns by $\pi/3$ with the same ratio of length.


How far from the center do we go?
If the question were the total distance this would be an infinite series. Because the question is the distance from the starting point it seems to me I somehow must use complex numbers and do an infinite sum of some kind. But not sure
 A: Short answer: take advantage of the periodicity: steps taken three turns apart are parallel.
With more detail: each step you take corresponds to adding a term of the form
\begin{equation}
\left(\frac{9}{10}\right)^ke^{ik\frac{\pi}{3}},
\end{equation}
starting from $k=0$.  The terminal point of our spiral is then
\begin{equation}
\sum_{k=0}^\infty \left(\frac{9}{10}\right)^ke^{ik\frac{\pi}{3}}\in\mathbb{C},
\end{equation}
assuming this series converges.  This seems an unwieldy modulus to compute at first, but we can take advantage of some periodicity.  Because $\exp(i(k+3)\frac{\pi}{3})=-\exp(ik\frac{\pi}{3})$ for every $k$, we have
\begin{align}
\sum_{k=0}^\infty \left(\frac{9}{10}\right)^ke^{ik\frac{\pi}{3}} &= \sum_{k=0}^\infty (-1)^k\left(\frac{9}{10}\right)^{3k}e^{0} + \sum_{k=0}^\infty (-1)^k\left(\frac{9}{10}\right)^{3k+1}e^{i\frac{\pi}{3}} + \sum_{k=0}^\infty (-1)^k\left(\frac{9}{10}\right)^{3k+2}e^{2i\frac{\pi}{3}}\\
&= \left(1 + \frac{9}{10}e^{i\frac{\pi}{3}} + \frac{81}{100}e^{2i\frac{\pi}{3}}\right)\sum_{k=0}^\infty(-1)^k\left(\frac{9^3}{10^3}\right)^k.
\end{align}
Because the alternating geometric series we see here will converge to
\begin{equation}
\sum_{k=0}^\infty(-1)^k\left(\frac{9^3}{10^3}\right)^k = \frac{1}{1+9^3/10^3},
\end{equation}
our spiral will converge in $\mathbb{C}$.  We can compute the modulus of the complex number that's left out front:
\begin{equation}
\bigg\vert1 + \frac{9}{10}e^{i\frac{\pi}{3}} + \frac{81}{100}e^{2i\frac{\pi}{3}}\bigg\vert = \frac{19\sqrt{91}}{100},
\end{equation}
meaning that the modulus of our terminal point is
\begin{equation}
\frac{1}{1+9^3/10^3}\frac{19\sqrt{91}}{100} = \frac{190\sqrt{91}}{1729}.
\end{equation}
A: There is an easier way to do this. As @AustinChristian correctly points out,
$$
\begin{equation}
z_n=\sum_{k=0}^n \left(\frac{9}{10}\right)^ke^{ik\frac{\pi}{3}}\in\mathbb{C},
\end{equation}
$$
But we also know that
$$
\begin{equation}
\sum_{k=0}^\infty p^k=\frac{1}{1-p},\quad |p|<1
\end{equation}
$$
Therefore, with $p=0.9e^{i\pi/3}$ we find that
$$\lim_{n\to \infty}z_n=\frac{1}{1-0.9e^{i\pi/3}}\approx 0.6044 + 0.8565i$$
This result is borne out by the figure of the spiral shown below.

