The close form of \(\displaystyle \sum_{k=0}^\infty z^k = \frac{1}{1 - z} \) for \(|z| < 1\) looks intuitively similar to that of the real counterpart \(\displaystyle \sum_{k=0}^\infty x^k = \frac{1}{1 - x} \) for \(x < 1\). However, for \(|z| = 1 \) and \( z \not = 1\), the value of the infinite series in the complex plane would keep walking on the unit circle !

Writing \(z = r \,e^{i \,\theta}\), on one hand, we can easily split the series into real and imaginary parts

\[ \begin{aligned} \sum_{k=0}^\infty z^k = \sum_{k=0}^\infty r^k \cos(k\,\theta) + i \sum_{k=0}^\infty r^k \sin(k\,\theta) \end{aligned} \]

On the other hand,

\[ \begin{aligned} \frac{1}{1-z} = \frac{1}{1 - r\,e^{i \,\theta}} = \cdots = \frac{1 - r \cos \theta + i r \sin \theta}{1 - 2r \cos \theta + r^2} \\ \end{aligned} \]

Check it! This necessarily means, in the peculiar case when \(|z| < 1\), ie when \(r < 1\),

\[ \begin{aligned} \sum_{k=0}^\infty r^k \cos(k \, \theta) = \frac{1 - r \cos \theta}{1 - 2r \cos \theta + r^2} \\ \sum_{k=0}^\infty r^k \sin(k \, \theta) = \frac{r \sin \theta}{1 - 2r \cos \theta + r^2} \\ \end{aligned} \]

!

More at Analysis of a Complex Kind.