# Branch points and branch cuts

1. Jan 12, 2012

### 1over137

Hello.

The following function $\Omega = [x^2 + y^2 + (z - ia)^2]^{1/2}$ has the following branch points:
$z_1 = i (a + \lambda), z_2 = i (a - \lambda)$ where $\lambda^2 = x^2 + y^2$.
Now I do not understand the theory about branch cuts. It is said in the paper I read: "We make cuts on the imaginary z axis from $z_1$ up to $z = i\infty$ and from $z_2$ down to $z = -i\infty$". OK. Now, let's consider the case $\lambda < a$. Then the lower branch line crosses the real z axis. What I do not understand is that
$\Omega^- = -\Omega^+ , z=0$
where $\Omega^-$ denotes the value of the radical for $z = 0^-$ and $\Omega^+$ for $z = 0^+$. (By the way, what is 'radical'?) They further wrote that
$\Omega^+ = [\lambda^2 + (-ia)^2]^{1/2} = -i(a^2-\lambda^2)^{1/2}$. How they calculated that? When I tried to calculate that I got +i instead of -i for the prefactor.
I thought that that lower branch cut means that the angle is from the interval $(-\pi/2,3/2\pi)$ and so
$\Omega^+ = [\lambda^2 + (\epsilon -ia)^2]^{1/2} = [\lambda^2 -a^2 + \epsilon^2 - 2\epsilon ai]^{1/2} = [a^2 - \lambda^2]^{1/2} e^{i\pi/2} = [a^2 - \lambda^2]^{1/2} i$ which is wrong.

Any help much appreciated.