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.