Order of branch point of z^(-1/2)

Oh okay. Having spent a rather long time reading about the difference between a pole and singularity, I found the most useful definition to be that a pole is a singularity that behaves likes the one ##\frac{1}{z^n}## has as ##z\to 0##. Since ##z^{1/2}## is not holomorphic at 0, and a branch point exits, this would exclude it from being a pole.
  • #1
247
11

Homework Statement


Give the location and orders of the branch points at, and classify the singularities [itex]f(z) =\frac{1}{z^{1/2}}[/itex]
Mod note: Fixed LaTeX in exponent above and below.
To OP: To write a fractional exponent such as ##z^{1/2}## use braces around the exponent, like this: ##z^{1/2}##. It works the same using the tex and itex tags.

Homework Equations




The Attempt at a Solution


My initial thought is there is a pole at z=0 of order 1/2? But I don't think you can have fractional order poles? So maybe I need to get the Laurent expansion for [itex]f(z) =\frac{1}{z^{1/2}}[/itex] about z=0? but I'm not too sure how to approach this?

Since we require z to 'go around' the complex plane twice to return to our original value is the branch cut of order 1?
I would assume that regardless of the order the branch cut extends along the real axis of the complex plane from 0 to infinity.

Many thanks :)
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Physgeek64 said:
Since we require z to 'go around' the complex plane twice to return to our original value
Where do we do that?

Check how the order of poles is defined.
 
  • #5
mfb said:
Where do we do that?

Check how the order of poles is defined.
As in, if you sub in ##e^{i2n\pi}## the 'n' that is required. I may be wrong, but this is what I was taught
 
  • #6
Physgeek64 said:
Ah okay, so it is an algebraic branch point of order 2?
Right. The singularity isn't a pole; it's a branch point. The term order refers to the number of Riemann sheets needed for the domain to make the function single-valued, not to the order of a pole, which doesn't apply here because you don't have a pole.
 
  • #7
vela said:
Right. The singularity isn't a pole; it's a branch point. The term order refers to the number of Riemann sheets needed for the domain to make the function single-valued, not to the order of a pole, which doesn't apply here because you don't have a pole.

That makes sense about the Riemann sheets. Sorry to be a pain- so even though the function is not analytic at z=0 there isn't a pole? I thought that a pole was simply any point in the complex plane that the function was ill-defined ? Thank you again, I appreciate the help
 
  • #8
Look up the definition of the term pole. A pole is a singularity, but not every singularity is a pole.
 
  • #9
vela said:
Look up the definition of the term pole. A pole is a singularity, but not every singularity is a pole.
I did- It said for a pole of the function f(z) at a, the function approaches infinity as z approaches a
 
  • #10
Is that the actual definition, or is it just a statement about how the function behaves if z=a is a pole?
 
  • #11
vela said:
Is that the actual definition, or is it just a statement about how the function behaves if z=a is a pole?
according to wikipedia https://en.wikipedia.org/wiki/Pole_(complex_analysis) however, I'm not sure of the extent to which this is accurate. On further exploration of the internet, another useful classification is that it must be possible to find an annulus around that point for which the function is analytic to be able to classify it as a point, which obviously excludes ##z^{-1/2}##. Does this sound somewhat reasonable, or not really?
 
  • #12
Wikipedia doesn't say, "If f(z) approaches infinity as ##z \to a##, then z=a is a pole." It says the converse: if z=a is a pole, then f(z) goes to infinity as z approaches a.

It defines a pole as a singularity that behaves likes the one ##\frac{1}{z^n}## has as ##z\to 0##.
 
Last edited:
  • #13
vela said:
Wikipedia doesn't say, "If f(z) approaches infinity as ##z \to a##, then z=a is a pole." It says the converse: if z=a is a pole, then f(z) goes to infinity as z approaches a.

It defines a pole as a singularity that behaves likes the one ##\frac{1}{z^n}## has as ##z\to 0##.
Oh okay. Having spent a rather long time reading about the difference between a pole and singularity, I found the most useful definition to be that ##z_0## is a pole if it is possible to define a function ##F(z)=1/f(z)## such that F is finite and holomorphic at ##z_0##. Since ##z^{1/2}## is not holomorphic at 0, and a branch point exits, this would exclude it from being a pole. Does this sound reasonable to you? Thank you :)
 
  • Like
Likes mfb

Suggested for: Order of branch point of z^(-1/2)

Back
Top