# contour integral with hyperbola branch cuts

I am reading Karl Graff's book on wave motion and stuck by a contour integral where the branch cuts are made of two hyperbolas in the 2nd and 4th quadrants. See figure attached Contour integral figure.

Here only the branch cut in the 2nd quadrant is shown as the only interest is in the upper half of the complex plane. The hyperbola degrades into a path along the positive imaginary axis and a section of the real axis on the negative side close to the origin after some manipulation as shown in the figure.

If two new polar coordinate systems are defined at the two branch points on the real axis, which are symmetric to the imaginary axis and define the ends of the cut on the real axis , as done in the book, the variable now takes the form of the product of coordinates redefined in the two new coordinate systems, say, $$\gamma$$ = $$\rho_1e^{i\theta_1}*\rho_2e^{i\theta_2}$$. In other words, the argument of $$\gamma$$ is $$\theta_1+\theta_2$$.

I can work out the argument by calculating the limit along each of the paths AO, OB, BO and ON but this is tedious. I am not able to read the argument out directly from the defined $$\theta_1$$ and $$\theta_2$$ as stated in the book. It seems to me something needs to be considered whereas the path is on one side of the complex plane when the coordinate is defined on the other. I do not know how this is done.

I would like to know how this works. Any advice that can help me improve my understanding on the technique, be it a book, a paper or a direct illustration, is appreciated.

## 1 Answer

I am here to answer my own question in case other people may face a similar problem. I have to acknowledge the help of a math Senior Lecturer in our university who solved this for me.

AO: $$\theta_1+\theta_2=\pi$$ (evident)

OB: $$\theta_1 (=\pi)+\theta_2(=0) = \pi$$

BO: $$\theta_1 (=\pi)$$ + $$\theta_2(=-2\pi$$, rotates clockwise one circle) = $$-\pi$$

OA: $$\theta_1 + \theta_2 = -\pi$$($$\theta_2$$ is negative)