I need to show that the following integral evaluates to

$$ I =\oint_{c}\frac{z^{1/2}}{1 + \,\sqrt{\, 2\,}\,z + z^{2}}\,\mathrm{d}z = 2^{3/2}\pi\mathrm{i}\sin\left(\frac{3 \pi}{8}\right) $$

Using Cauchy's Residue Theorem, there $c$ is the keyhole contour and the branch cuts have been chosen so that the a number in the complex plane can be written as $r\mathrm{e}^{\mathrm{i}\theta}$ where $-\pi < \theta < \pi$.

I know that I need to convert the function inside the integral into a function of the form

$$ \mathrm{f}\left(z\right) = \frac{\phi\left(z\right)} {\left(z - z_{0}\right)^{m}} $$

so that I can find out where the poles are which then allows me to find the residues, but Im just having some trouble massaging it into the correct format.

Anybody have any tips or suggestions ?.


1 Answer 1


The function $\frac{z^{1/2}}{z^2+\sqrt2 z+1}$ has a branch point at $z=0$ and poles at $\frac{-\sqrt 2\pm i\sqrt{2}}{2}=e^{\pm i3\pi/4}$. Therefore with $C$ as the classical keyhole contour with the keyhole along the negative real axis we have

$$\begin{align} \oint_C \frac{z^{1/2}}{z^2+\sqrt2 z+1}\,dz&=2\pi i \text{Res}\left(\frac{z^{1/2}}{z^2+\sqrt2 z+1}, z=\frac{-\sqrt 2\pm i\sqrt{2}}{2}\right)\\\\ &=2\pi i \left( \frac{e^{i3\pi/8}}{2i\sin(3\pi/4)}+\frac{e^{-i3\pi/8}}{-2i\sin(3\pi/4)}\right)\\\\ &=2\pi i \frac{\sin(3\pi/8)}{\sin(3\pi/4)}\\\\ &=2^{3/2}\pi i \sin(3\pi/8) \end{align}$$

as was to be shown.


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