So we all know that the continued fraction containing all $1$s...

$$ x = 1 + \frac{1}{1 + \frac{1}{1 + \ldots}} $$

yields the golden ratio $x = \phi$, which can easily be proven by rewriting it as $x = 1 + \dfrac{1}{x}$, solving the resulting quadratic equation and assuming that a continued fraction that only contains additions will give a positive number.

Now, a friend asked me what would happen if we replaced all additions with subtractions:

$$ x = 1 - \frac{1}{1 - \frac{1}{1 - \ldots}} $$

I thought "oh cool, I know how to solve this...":

\begin{align} x &= 1 - \frac{1}{x} \\ x^2 - x + 1 &= 0 \end{align}

And voila, I get...

$$ x \in \{e^{i\pi/3}, e^{-i\pi/3} \} $$

Ummm... why does a continued fraction containing only $1$s, subtraction and division result in one of two complex (as opposed to real) numbers?

(I have a feeling this is something like the $\sum_i (-1)^i$ thing, that the infinite continued fraction isn't well-defined unless we can express it as the limit of a converging series, because the truncated fractions $1 - \frac{1}{1-1}$ etc. aren't well-defined, but I thought I'd ask for a well-founded answer. Even if this is the case, do the two complex numbers have any "meaning"?)