I don't understand the following argument in my lecture notes:
Consider the following ODE: $$ w''(z) + f(z) w'(z) + g(z) w(z) = 0, \quad z \in \mathbb{C}$$ where $$f(z) = f_0 + \frac{f_1}{z} + \frac{f_2}{z^2} + \ldots$$ $$ g(z) = g_0 + \frac{g_1}{z} + \frac{g_2}{z^2} + \ldots, $$ where $f_0$, $g_0$ and $g_1$ are not all zero. Trying the ansatz that $w_j (z) \sim e^{\lambda z} z^{ \mu_j}$, $j=1,2$ and plugging into the ODE gives $\lambda_1 > \lambda_2$ being roots of the equation $\lambda^2 + f_0 \lambda + g_0 = 0$, and $\mu_j = - \frac{f_1 \lambda_j + g_1}{2 \lambda_j + f_0}$, for $j=1,2$. Then it mentions that by considering branch cuts, $$w_1 (z) \sim e^{\lambda_1 z} z^{\mu_1}, \quad \quad \text{ for Arg}(z) \in ( -\frac{3 \pi}{2}, \frac{3 \pi}{2}).$$ Therefore, $$w_1 (z e^{2 \pi i}) \sim e^{\lambda_1 z} z^{\mu_1} e^{2 \pi i \mu_1}, \quad \quad \text{ for Arg}(z) \in ( -\frac{7 \pi}{2}, -\frac{ \pi}{2}).$$ Since $w_1$ and $w_2$ are linearly independent solutions, there exists a constant $K_{1,2}$ such that $$ w_1 ( z e^{2 \pi i}) = e^{2 \pi i \mu_1} w_1 (z) + K_{1,2} w_2 (z).$$
I am really perplexed about why we are considering $\text{Arg}(z) \in ( -\frac{3 \pi}{2}, \frac{3 \pi}{2})$ in the first place. Where does the value $\frac{3 \pi}{2}$ come from? Any suggestions or comments?