Branch cut of $\sqrt{z}$ along the *positive* real axis

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...

• 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 $0<\theta<2\pi$). – MPW Dec 29 '14 at 12:41
• But if I take $(0+\epsilon)$ and $(2\pi - \epsilon)$ letting $\epsilon \rightarrow 0$ I get the same limits...? – SuperCiocia Dec 29 '14 at 13:06
• The discontinuity will be in the real part... – Hans Lundmark Dec 29 '14 at 13:23

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
$$\lim_{\epsilon \to 0^+} \cos \Big(\frac{\epsilon}{2}\Big) \neq \lim_{\epsilon \to 0^-} \cos \Big(\frac{2\pi - \epsilon}{2}\Big)$$
• 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)? – SuperCiocia Dec 29 '14 at 14:03
• 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? – SuperCiocia Dec 29 '14 at 14:10
• 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. – Aaron Maroja Dec 29 '14 at 14:28