square root of complex number in a resolvent operator Let $ R_\lambda (x,y) = \frac{1}{2\sqrt{-\lambda}} e^{-\sqrt{-\lambda}|x-y|}$ . 
Now according to the book I am following if $\lambda > 0$ and we shift $\lambda \rightarrow \lambda + i\epsilon$, we get
$ R_\lambda (x,y) = \frac{1}{2\sqrt{-\lambda}} e^{-\sqrt{-\lambda}|x-y|} \rightarrow \frac{i}{2\sqrt{\lambda}}e^{i\sqrt{\lambda}|x-y| -\epsilon|x-y|/2\sqrt{\lambda}}$
To show this I had $\sqrt{-\lambda}=\sqrt{|\lambda|}e^{-i\pi/2} = -i \sqrt{|\lambda|}$, and then I expanded $\sqrt{-\lambda -i\epsilon} = -i\sqrt{|\lambda|} + \frac{i\epsilon}{-2i\sqrt{\lambda}}$. However, this gives a positive sign for the $\epsilon$ term of the exponential in the resolvent. 
Is there anything I am missing here? Any help will be appreciated.
 A: I guess it all depends whether the phase of your complex number is allowed to "temporarily" go outside $(-\pi, \pi]$ interval.  $\lambda + i \epsilon = \sqrt{\lambda^2 + \epsilon^2} e^{i \delta}$, where $\delta = tan^{-1} \frac{\epsilon}{\lambda}$, i.e. $\delta$ is small positive quantity. Now $-1 = e^{i \pi}$ and $-\lambda - i \epsilon = \vert \lambda \vert e^{i (\pi + \delta)}$.  Notice that $\pi + \delta$ jumps outside the interval, and you can choose to reduce it back by subtracting $2 \pi$. Let's not that. Then $\sqrt{-(\lambda + i \epsilon)} = \sqrt{\vert \lambda \vert } e^{i \frac{\pi}{2} + i \frac{\delta}{2}} \approx \sqrt{\vert \lambda \vert } ( i - \sin \frac{\delta}{2} )$, so it would work as the book says. 
If you choose, however, to perform the modular reduction, than $-\lambda - i \epsilon = \vert \lambda \vert e^{i (-\pi + \delta)}$ and hence $\sqrt{-(\lambda + i \epsilon)} = \sqrt{\vert \lambda \vert } e^{-i \frac{\pi}{2} + i \frac{\delta}{2}} \approx \sqrt{\vert \lambda \vert } ( -i + \sin \frac{\delta}{2} )$, which is the result you obtained.
