1. ## Residue

I have a doubt with the singularities and poles of a function, for example in the next complex function:

$\displaystyle f(x)=\sqrt{x^2+a}$

with $\displaystyle a$ a constant. The function is zero when $\displaystyle x=\pm{\sqrt{a}}$

For this value of x, is it a singularity or a pole? If it is a pole, of which order? Is it possible to calculate the residue for $\displaystyle x=\pm{\sqrt{a}}$?

2. Originally Posted by germana2006
I have a doubt with the singularities and poles of a function, for example in the next complex function:

$\displaystyle f(x)=\sqrt{x^2+a}$

with $\displaystyle a$ a constant. The function is zero when $\displaystyle x=\pm{\sqrt{a}}$ Mr F says: No. You want $\displaystyle {\color{red}x=\pm i {\sqrt{a}}}$.

For this value of x, is it a singularity or a pole? If it is a pole, of which order? Is it possible to calculate the residue for $\displaystyle x=\pm{\sqrt{a}}$?
It's a branch point. A branch point is a special type of singularity.

3. ## This may help

$\displaystyle f(x)=\sqrt{x^2+a}=e^{ln(\sqrt{(x^2+a)})}=e^{1/2ln(x^2+a)}$

4. Originally Posted by mr fantastic
It's a branch point. A branch point is a special type of singularity.
I disagree. To be a singularity we require that the function is analytic in a puntured neighborhood of that point. This is not true for $\displaystyle \sqrt{z^2+a}$ along its branch.

5. Originally Posted by ThePerfectHacker
I disagree. To be a singularity we require that the function is analytic in a puntured neighborhood of that point. This is not true for $\displaystyle \sqrt{z^2+a}$ along its branch.
I quote from Churchill et al Complex Variables and Applications:

"... a point $\displaystyle z_0$ is called a singular point of a function f if f faails to be analytic at $\displaystyle z_0$ but is analytic at some point in every neighbourhood of $\displaystyle z_0$. A singular point is said to be isolated if, in addition, there is some neighbourhood of $\displaystyle z_0$ throughout which f is analytic except at the point itself."

A branch point is a singular point - what it's not is an isolated singular point.

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There are two categories of singularity:

1. Isolated singularity.

The point $\displaystyle z = z_0$ is an isolated singularity of $\displaystyle f(z)$ if $\displaystyle f(z)$ is not analytic at $\displaystyle z = z_0$ but is analytic in some deleted neighbourhood of $\displaystyle z = z_0$.

There are three types:

i. Removable. $\displaystyle \lim_{z \rightarrow z_0} f(z)$ exists and is finite. Alternatively, the Laurent series expansion of $\displaystyle f(z)$ about $\displaystyle z = z_0$ has no principle part.

Eg. $\displaystyle f(z) = \frac{\sin z}{z}$ has a removable singularity at z = 0.

ii. Pole. $\displaystyle (z - z_0)^m f(z)$ is analytic at $\displaystyle z = z_0$ for some positive integer m. The smallest such m is called the order of the pole. Alternatively, the principle part of the Laurent series expansion of $\displaystyle f(z)$ about $\displaystyle z = z_0$ terminates after m terms.

Eg. $\displaystyle f(z) = \frac{1}{z-1}$ has a simple pole at z = 1.

iii. Essential. It's neither removable nor a pole. Alternatively, the principle part of the Laurent series expansion of $\displaystyle f(z)$ about $\displaystyle z = z_0$ does not terminate. Alternatively, $\displaystyle \lim_{z \rightarrow z_0} f(z)$ does not exist (NB. Picard's Theorem).

Eg. $\displaystyle f(z) = e^{1/z}$ has an essential singularity at z = 0.

2. Non-isolated singularity.

If the singular point $\displaystyle z = z_0$ is not isolated, then it is a non-isolated singularity.

Eg. The branch point z = 0 of $\displaystyle f(z) = \ln z$ is a non-isolated singularity. In fact, any branch point of a multiple-valued function is a non-isolated singularity.

6. Okay. I just never seen such terms before. But if you use that terminology then it is not just $\displaystyle \pm i\sqrt{a}$ it then should be $\displaystyle (-\infty,0]$.

7. Originally Posted by ThePerfectHacker
Okay. I just never seen such terms before. But if you use that terminology then it is not just $\displaystyle \pm i\sqrt{a}$ it then should be $\displaystyle (-\infty,0]$.
Yes. But it should be noted that branch cuts are defined by convention ..... You say $\displaystyle (-\infty,0]$, I could just as well define the branch cut to be $\displaystyle [0, \infty)$ in which case the points in $\displaystyle (-\infty,0]$ are not singular .....

The branch point is (the only point that is) common to all branch cuts .... This makes it fundamentally different from all other points on a branch cut.