How can I take the complex logarithm of this equation?

Question

I have $$ e^{i\beta}=\pm\frac{g_1}{\sqrt{1-g_0^2}} $$ where $g_0$ is a real value and $g_1$ is complex. I'm trying to solve for $\beta$ in the equation, but I'm not quite sure how I can do that. I tried to take the logarithm so that $$ \beta =-i \ln \left(\pm\frac{g_1}{\sqrt{1-g_0^2}}\right) $$ However, the answer looks like $$ \beta'=-i\log\frac{g_1}{\sqrt{1-g_0^2}}+k\pi,\ \text{where}\ k\in\mathbb{Z} $$ I'm wondering where does the $\pm$ sign go? Why there's a $+k\pi$ term? Is that relevant to the Riemann surface? Thanks:)

Answer 1

Since $e^{i \theta} = e^{i(\theta + 2k\pi)}$ for any $k \in \mathbb{Z}$, when you solve an equation such as $$ e^{i \theta} = w $$ for $\theta \in \mathbb{C}$, you get one value for each integer. (The complex logarithm is multivalued.) However, in your equation, since you have the $\pm$ sign, you get twice as many solutions! To wit, $$ e^{i(\theta + (2k+1)\pi)} = e^{i(\theta + 2k\pi) + i\pi} = e^{i(\theta + 2k\pi)} e^{i\pi} = -e^{i(\theta + 2k\pi)}. $$

Therefore, when you're solving an equation such as $e^{i\theta} = \pm w$, you get both the even and odd multiples of $\pi$, i.e. all the integer multiples of $\pi$.