Science > Physics > Questions about branch cuts and residue calculus...
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Science > Physics |
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"" |
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25 Jul 2007 11:25:13 AM |
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Questions about branch cuts and residue calculus... |
Hi all,
For this complex-valued function:
F(z)=log(z*f(z)/g(z))
where z=0 is a removable singularity, and the z0 such that g(z0)=0 is
a non-removable singularity.
My questions are:
1. Is z=0 still a branch point and hence it induces a branch cut?
2. Is z0 such that g(z0)=0 a branch point inducing a branch cut?
3. I am looking for calculus rules about residues, for example, change-
of-variable rule,
if z=h(y) where y is another complex variable.
Knowing the residue F(z) with respect to z, at z=z0, will it be able
to give me the residue of y for F(h(y)) at y=y0, such that
g(h(y0))=0?
In general, I am interested in learning what calculus and algebraic
rules are avaiable from the residue world.
And further, if I have FG(z) = F(z) * G(z) , knowing G(z)'s residue at
z=z0, and z=z0 is a non-removable singularity for G(z) but not for
F(z), will I be able to compute the residue of FG(z) at z=z0?
.
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| User: "Robert Israel" |
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| Title: Re: Questions about branch cuts and residue calculus... |
25 Jul 2007 12:53:20 PM |
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writes:
Hi all,
For this complex-valued function:
F(z)=log(z*f(z)/g(z))
where z=0 is a removable singularity, and the z0 such that g(z0)=0 is
a non-removable singularity.
My questions are:
1. Is z=0 still a branch point and hence it induces a branch cut?
If by "z=0 is a removable singularity" you mean it's a removable singularity
of F(z), then by definition it is not a branch point and there is no branch
cut there: the function F(z) must be analytic in {z: 0 < |z| < r} for some
r > 0, and has a finite limit as z -> 0. This will happen (for a suitable
branch of log) if z*f(z)/g(z) has a finite nonzero limit as z -> 0.
2. Is z0 such that g(z0)=0 a branch point inducing a branch cut?
Assuming z0*f(z0) is nonzero and f and g are analytic near z0, yes.
3. I am looking for calculus rules about residues, for example, change-
of-variable rule,
if z=h(y) where y is another complex variable.
Knowing the residue F(z) with respect to z, at z=z0, will it be able
to give me the residue of y for F(h(y)) at y=y0, such that
g(h(y0))=0?
I assume this F(z) is not the same as the one in question 2, because residues
are only defined for isolated singularities, not branch points.
The residue of F(z) at z=z0 is the coefficient of (z-z0)^(-1) in the
Laurent series of F(z) around z=z0. Now if F(z) has a simple pole
at z=z0 with residue r and h(y) - z0 has a simple zero at y0, we have
F(z) = r/(z-z0) + ... and h(y) = z0 + h'(y0) (y - y0) + ...
so F(h(y)) = r/(h'(y0)(y - y0) + ...) + ... = r/(h'(y0) (y-y0)) + ....
Thus the residue of F(h(y)) at y=y0 will be r/h'(y0). However, things
will be more complicated if you don't have a simple pole and a simple zero.
In general, I am interested in learning what calculus and algebraic
rules are avaiable from the residue world.
And further, if I have FG(z) = F(z) * G(z) , knowing G(z)'s residue at
z=z0, and z=z0 is a non-removable singularity for G(z) but not for
F(z), will I be able to compute the residue of FG(z) at z=z0?
Again, if G has a simple pole at z=z0, while F is analytic there (if it has a
removable singularity there, then remove it), the answer is yes: the residue
of FG will be r*F(z0) if the residue of G is r. If it's not a simple pole,
the Laurent coefficients for other negative powers of z-z0 will have an effect.
--
Robert Israel
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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