Start with the geometric series: \[ \sum_{n=0}^\infty z^n = 1 + z + z^2 + z^3 + \cdots\text. \] If $z$ is not a real number, the partial sums of this series form a spiral pattern, which appears to be self-similar. Below is the pattern for $z = (1+i)/2$. (Click for interactive graph.)
A similarity of the plane can be expressed as a (complex) affine function $f(w) = aw + b$. To determine $a$ and $b$ for our spiral, we note that $f$ should satisfy \[ f(0) = 1 \qquad\text{and}\qquad f(1) = 1+z\text. \] The first equation implies $b = 1$. From the second equation, we then have $a + 1 = 1 + z$, so $a = z$. Our function is therefore $f(w) = zw + 1$, with $z$ treated as a constant.
Now we check whether this function $f(w)$ does in fact show that the spiral is self-similar. Indeed, $f(1+z) = z(1+z) + 1 = 1 + z + z^2$, and more generally, the sequence $1$, $f(1)$, $f(f(1))$, $f(f(f(1)))$, … coincides with the sequence of partial sums of the geometric series.
If $z \ne 1$, then the function $f(w)$ has a fixed point. Solve $f(w) = w$, or $zw + 1 = w$, to find that the fixed point is $w = 1/(1 - z)$. When $|z| < 1$, this fixed point is the sum of the series. But in general it has a nice geometric connection to the series, even when the series diverges: it is the center of similarity for the sequence of partial sums. For example, when $z = i+1/\sqrt3$, the partial sums and center look like this:
In particular, when $z$ is a root of unity, the partial sums of the series lie at the vertices of a regular polygon, and $1/(1-z)$ is the center of this polygon. (It is also the Cesàro sum of the series in that case.) Here are the polygons and centers for $z = e^{i\pi/4}$ and $z = e^{i\,3\pi/5}$:
When $z = -1$, the polygon collapes to a line segment, and of course the center of this segment is $1/2$, the usual value assigned to the series $1 - 1 + 1 - 1 + \cdots$.
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