I really don't understand how we determine branches for root. For log I know that:

$log(z)= ln(r)+i(\theta + 2pin)$ where n=..-1,0,1,... which shows multivalued

\[
\theta = \pi + \frac{\pi}{2} = \frac{3\pi}{2}, \quad (\cos(\frac{3\pi}{2}), \sin(\frac{3\pi}{2})) = (0, -1).
\]

\[
\theta = \pi - \frac{\pi}{2} = \frac{\pi}{2}, \quad (\cos(\frac{\pi}{2}), \sin(\frac{\pi}{2})) = (0, 1).
\]

\[
\theta = \pi \pm \frac{\pi}{2} \implies
\begin{cases}
\theta = \frac{\pi}{2}, & (\cos(\frac{\pi}{2}), \sin(\frac{\pi}{2})) = (0, 1), \\
\theta = \frac{3\pi}{2}, & (\cos(\frac{3\pi}{2}), \sin(\frac{3\pi}{2})) = (0, -1).
\end{cases}
\]

$$
\theta\left(\frac{\pi}{2}\right) = 0
$$
$$
\text{If I make a 180° turn, I arrive at } \pi
$$
$$
\int_0^\pi f(\theta) \, d\theta
$$

$$
\theta = 0 \text{ at } \pi
$$
$$
\text{Half path above: } \theta = -\frac{\pi}{2}
$$
$$
\text{Half path below: } \theta = \frac{\pi}{2}
$$
$$
\text{Counterclockwise direction: } \theta \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]
$$

$$
\theta = 0 \text{ at } \frac{\pi}{2}
$$
$$
\text{If I make half a turn counterclockwise, I arrive at } \pi
$$
$$
\text{The path is from } 0 \text{ to } \pi
$$
$$
\text{Counterclockwise direction: } \theta \in [0, \pi]
$$

$$
\sum_{n=1}^\infty \ln(27n^3 - 2n) - 3 \ln(3n).
$$

$\ln\bigl(27n^3 - 2n\bigr) \;=\; \ln\bigl[n\,(27n^2 - 2)\bigr]
\;=\; \ln(n) \;+\; \ln\bigl(27n^2 - 2\bigr)$

$\ln(27n^3 - 2n)\;-\;3\,\ln(3n)
\;=\;\bigl[\ln(n)+\ln(27n^2 - 2)\bigr]
\;-\;\ln(27n^3). $

$\ln(n)\;+\;\ln(27n^2 - 2)\;-\;\ln(27n^3)
\;=\;\ln\!\Bigl[\tfrac{n\,(27n^2 - 2)}{27n^3}\Bigr]
\;=\;\ln\!\Bigl[\tfrac{27n^2 - 2}{27n^2}\Bigr].$

$\tfrac{27n^2 - 2}{27n^2}
\;=\;1 \;-\;\tfrac{2}{27n^2}$

$\ln\bigl(27n^3 - 2n\bigr)\;-\;3\,\ln(3n)
\;=\;
\ln\!\Bigl(1 - \tfrac{2}{27\,n^2}\Bigr)$

$\sum_{n=1}^\infty \ln\!\Bigl(1 - \tfrac{2}{27n^2}\Bigr)
\;=\;
\ln\!\Bigl[\prod_{n=1}^\infty \Bigl(1 - \tfrac{2}{27n^2}\Bigr)\Bigr]$

$\sin(\pi x)
\;=\;
\pi\,x \,\prod_{n=1}^\infty
\Bigl(1 - \tfrac{x^2}{n^2}\Bigr)$

$\sum_{n=1}^\infty
\Bigl[\ln(27n^3 - 2n) - 3 \ln(3n)\Bigr]
\;=\;
\ln\!\Bigl[\prod_{n=1}^\infty \Bigl(1 - \tfrac{2}{27n^2}\Bigr)\Bigr]
\;=\;
\ln\!\Bigl[
\frac{\sin\bigl(\pi \sqrt{2/27}\bigr)}{\pi \,\sqrt{2/27}}
\Bigr]$