3
$\begingroup$

I am a physicist who usually doesn't need to care about the fact that square root is not single-valued on the complex plane. But I would like to give a meaning to and compute the contour integrals :

\begin{align} \oint_w \frac{z-a}{(z-w)^{1/2}}dz \end{align} and \begin{align} \oint_w \frac{z-a}{(z-w)^{3/2}}dz, \end{align} where $a$ is a complex number and the contour is a small circle around $w$. I have some very basic knowledge of complex analysis, know how to use residues, but I am not familiar with contour integrals which branch cuts. I don't find people considering these integrals online, so I would like to know if it is possible to give them a meaning and to compute them.

For the first integral, if the variable is changed so that the contour is around 0, there is a branch cut from $-\infty$ to 0. For the second integral there would be two branch points, $w$ and $-w$. I can deform the contours to avoid branch cuts, but how can I then compute the different integrals?

Thank you for your help.

$\endgroup$
6
$\begingroup$

To evaluate the contour integrals, you can simply parameterize. For the first example, set $z=w+r e^{i \phi}$, $\phi \in [-\pi,\pi]$. Then the integral is equal to

$$r^{1/2} \int_{-\pi}^{\pi} d\phi \, e^{i \phi/2} \left (w-a+r e^{i \phi}\right ) = 4 (w-a) r^{1/2} - \frac{4}{3} r^{3/2} $$

The other integral may be done similarly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.