How can we show this complex exponential identity? Let $\kappa\ge0$, $y_\pm:=\sqrt\kappa\pm\sqrt{\kappa+4}$ and $$H(w):=\frac{2y_+\ln(w-y_-)-2y_-\ln(w-y_+)}{y_+-y_-}\;\;\;\text{for }w\in\mathbb C.$$
How do we obtain the identity $$B:=\exp\left(\frac12H\left(2\sqrt\kappa\right)\right)=2\left(-\frac{y_+}{y_-}\right)^{\frac12\frac{\sqrt\kappa}{\sqrt{\kappa+4}}}\exp\left(\frac12\pi{\rm i}\left(1-\frac{\sqrt\kappa}{\sqrt{\kappa+4}}\right)\right)\tag1$$ (which can be found in this paper in eq. 35)?
What I get is $$H\left(2\sqrt\kappa\right)=\frac{y_+\ln(y_+)-y_-\ln(y_-)}{\sqrt{\kappa+4}}=\frac{\sqrt\kappa}{\sqrt{\kappa+4}}(\ln(y_+)-\ln(y_-))+\ln(y_+)+\ln(y_-)\tag2$$ and hence $$B=y_+^{\frac{y_+}{2\sqrt{\kappa+4}}}y_-^{-\frac{y_-}{2\sqrt{\kappa+4}}}\tag3,$$ but I don't know how to further proceed ...
 A: We have, as long as we choose the principal branch of the complex logarithm and attribute to negative real numbers $s$ the value $\ln(s) := \ln(-s) + i\pi$, and knowing that $-y_+y_- = \kappa + 4 - \kappa = 4$ :
$$\begin{split} B &= \exp\left(\frac{1}{2}\left(\frac{\sqrt\kappa}{\sqrt{\kappa+4}}\big(\ln(y_+)-\ln(y_-)\big)+\ln(y_+)+\ln(y_-)\right)\right)\\
& = \exp\left(\frac{1}{2}\left(\frac{\sqrt\kappa}{\sqrt{\kappa+4}}\Big(\color{blue}{\ln(y_+)-\big(\ln(-y_-)} + \color{green}{\pi i}\big)\Big)+\color{red}{\ln(y_+)+\big(\ln(-y_-)} + \color{green}{\pi i}\big)\right)\right)\\
& = \color{blue}{\exp\left(\frac{1}{2}\cdot\frac{\sqrt\kappa}{\sqrt{\kappa+4}}\ln\left(-\frac{y_+}{y_-}\right)\right)}\color{red}{\exp\left(\frac{1}{2}\ln(-y_+y_-)\right)} \color{green}{\exp\left(- \frac{1}{2}\cdot\frac{\sqrt\kappa}{\sqrt{\kappa+4}}\pi i + \frac{1}{2}\pi i\big)\Big)\right)}\\
& = \color{blue}{\left(-\frac{y_+}{y_-}\right)^{\frac12\frac{\sqrt\kappa}{\sqrt{\kappa+4}}}} \cdot \color{red}{2} \cdot \color{green}{\exp\left(\frac12\pi{\rm i}\left(1-\frac{\sqrt\kappa}{\sqrt{\kappa+4}}\right)\right)}\end{split}$$
