How to find the branch cuts of $\sqrt{g(z)}$ and the contour integral $\int_{z_1}^{z_2}d z\sqrt{g(z)}$

Question

I need to evaluate the following integral: \begin{equation} \int_{z_1}^{z_2} d z \sqrt{g(z)}, \end{equation} where the function $g(z)$ is given by \begin{equation} g(z)=-\left(\alpha-\frac{\beta}{z^2}+\frac{\gamma}{z^4}\right)=\frac{\alpha\,(z^2-z_1^2)(z_2^2-z^2)}{z^4}, \end{equation} and $z_1$ and $z_2$ are two zeros of $g(z)$ with positive real parts.

When $z_1$ and $z_2$ are real numbers, by the change of variable $z^2=z_2^2-(z_2^2-z_1^2)\sin^2(\theta)$ I write the integral in the form of elliptic functions. The result is:
\begin{equation} 2\sqrt{\alpha} z_2 \left((1-\frac{m^2}{2})K(m)-E(m)\right), \end{equation} where $m:=(z_2^2-z_!^2)/z_2^2$ and $K(m)$ and $E(m)$ are respectively the complete elliptic integrals of the first and second kind.

The problem arises when $z_1=z_2^*$ are complex conjugate roots. In this case, in the above integral the path lies along the imaginary axis and the branch of $\sqrt{g(z)}$ is real and positive. I don't know how to perform the integration in this case. But, surprisingly, for some values of $z_1=z_2^*$, the numerical value obtained by Mathematica is exactly identical with the result obtained in the case of real zeros of $g(z)$.

My questions are:

(1) Which appropriate contour I should take to perform the integral in the case $z_1=z_2^*$?

(2) Is there any analytical result for the integral in this case? If yes, how can I find it?

(3) Two special choices of the parameters may be helpful. $(\alpha, \beta, \gamma)=(1.5, 0.75, 0.25)$ and $(\alpha, \beta, \gamma)=(4, -0.05, 0.1225)$.

Any help will be appreciated.

Answer 1

Let $\alpha = 1$. I assume that you mean that it's possible to choose a branch of $\sqrt {g(z)}$ in such a way that $I = \int_{z_1}^{z_2} \sqrt {g(z)} \, dz$ is real when $z_1 = z_2^*$; $\sqrt {g(z)}$ itself isn't real on $[z_1, z_2]$.

Since the possible values of $m$ are such that $|m - 1| = 1 \land m \neq 0$ (in particular, $m = 2$ when $\arg z_2 = \pi/4$), we need to construct a branch of $$\left( 1 - \frac m 2 \right) K(m) - E(m)$$ (not $1 - m^2/2$) which is continuous at those points. For this, we can use the algebraic relation between $E$ and $K$ and take $$K(m) = m^{-1/2} (K(m^{-1}) + i K(1 - m^{-1})),$$ then the arguments do not lie on $[1, \infty)$. We obtain $$I = z_2 \frac {(2 E(1 - m) - m K(1 - m)) (K(m^{-1}) + i K(1 - m^{-1})) - \pi m^{1/2}} {m^{1/2} K(1 - m)}, \\ m = 1 - \left( \frac {z_2^*} {z_2} \right)^{\! 2}, \quad 0 < \arg z_2 < \frac \pi 2,$$ where $E(m)$ and $K(m)$ are analytic continuations from $(-\infty, 1)$ to $\mathbb C \setminus [1, \infty)$, which I assume is what is desired for numerical calculations.