1
$\begingroup$

Consider the function $\sqrt{z}$, with $z\in\mathbb{C}$.

Writing $z = re^{i\theta}$, the imaginary part of $\sqrt{z}$ can be expressed as: $$Im({\sqrt{z}) = r^{\frac{1}{2}}\sin({\frac{\theta}{2} + k\pi})}$$

with $k\in\mathbb{Z}$. For the time beind let's set $k=0$ anyway.

Now, if I choose the principal value of the function to be $-\pi<\theta<\pi$, I can cleary see that $Im(\sqrt{z})$ has a branch cut along the negative real axis because $$\sin(\pi/2) \neq \sin(-\pi/2)$$.

Now: I want the branch cut to be along the positive real axis. I would set the principal value to be $0<\theta<2\pi$, but then, along said axis, $\sin(0/2) = \sin(2\pi/2) = 0$ so there is no discontinuity...

$\endgroup$
  • 2
    $\begingroup$ If you make a cut along the positive real axis, you are omitting the points in the cut from the domain. You are excluding $\theta=0$ (remeber, you require $0<\theta<2\pi$). $\endgroup$ – MPW Dec 29 '14 at 12:41
  • $\begingroup$ But if I take $(0+\epsilon)$ and $(2\pi - \epsilon)$ letting $\epsilon \rightarrow 0$ I get the same limits...? $\endgroup$ – SuperCiocia Dec 29 '14 at 13:06
  • 1
    $\begingroup$ The discontinuity will be in the real part... $\endgroup$ – Hans Lundmark Dec 29 '14 at 13:23
1
$\begingroup$

A branch cut is a curve or a line that is introduced in order to define a branch $F$ of a multivalued funtion, in this case $f(z) = \sqrt{z}$. Now points of a branch cut are singular points of $F$. You want to be along the positive real axis, the origin and the ray $\theta = 0 $ will make up the branch cut.

Notice that the discontinuity will be if we take $\sqrt{z} = \sqrt{r}e^{i\frac{\theta}{2}} = \sqrt{r}\Bigg(\cos \Big(\frac{\theta}{2}\Big) + i\sin\Big(\frac{\theta}{2}\Big)\Bigg)$ , where $0 < \theta < 2\pi$. Then

$$\lim_{\epsilon \to 0^+} \cos \Big(\frac{\epsilon}{2}\Big) \neq \lim_{\epsilon \to 0^-} \cos \Big(\frac{2\pi - \epsilon}{2}\Big) $$

$\endgroup$
  • $\begingroup$ Thanks. So if I had to take the integral of $\frac{1}{\sqrt{z^2+m^2}}$ like this question (physics.stackexchange.com/questions/94074/…), which principal value should I choose in order for the branch cut to be on the imaginary axis (as shown in the diagram)? $\endgroup$ – SuperCiocia Dec 29 '14 at 14:03
  • $\begingroup$ Find the poles and choose a branch cut to work with. $\endgroup$ – Aaron Maroja Dec 29 '14 at 14:04
  • $\begingroup$ But the poles ($+im$ and $-im$) do not have a Laurent expansion because of the branch cut. I don't know how to choose the branch cut. Does it make sense it's on the $y$ axis from $im$ to $\infty$? Why? $\endgroup$ – SuperCiocia Dec 29 '14 at 14:10
  • $\begingroup$ This is a different question, you know. But it makes sense taking the branch cut the line $iy$, where $y \geq m$ so you can work the contour integral by finding a parametrization to $C_1$ and fincding the residue to apply Cauchy's Theorem. Adam has given a nice answer. $\endgroup$ – Aaron Maroja Dec 29 '14 at 14:28

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.