Contour Integral over unit circle

May 2010
43
1
I have attempted to solve the following contour integral over the complex unit circle, subject to the constraint that f(z) be analytic over the entire area bounded by the circle, and |a| < 1. Can someone tell me if the following is correct:

\(\displaystyle \oint f(z)dz/(z^n-a) = \frac{2\pi i}{na^{((n-1)/n)}}\sum_{k=0}^{n-1}f(\zeta _{k})exp^{2\pi ik(n-1)/n}\)

where: \(\displaystyle \zeta_{k}=a^{1/n}exp^{2\pi ik/n}\) and k=0,1,2,...n-1

If true, this would be very interesting, at least to me, because it means a Fourier transform (or something very like one) is really just the sum of the residues of a function evaluated at the complex roots of \(\displaystyle a^{1/n}\)
 
May 2010
274
67
Los Angeles, California
It might be correct. Did you do some simplification here? We have that:

\(\displaystyle \frac{f(z)}{z^n-a}=\frac{f(z)}{(z-\zeta_0)\cdots (z-\zeta_{n-1})}\).

After applying Cauchy's Residue Theorem, you need to simplify expressions of the form:

\(\displaystyle (\zeta_k-\zeta_0)(\zeta_k-\zeta_1)\cdots \widehat{(\zeta_k-\zeta_k)}\cdots (\zeta_k-\zeta_{n-1})\), where \(\displaystyle \widehat {\phantom{ooo}}\) means omitted. Is this what you did?
 
May 2010
43
1
I used \(\displaystyle \sum_{k=0}^{n-1}\lim_{z\to \zeta _k}(z-\zeta _k)f(\zeta _k)/(z^n - 1/a)\) for the residues and L'hopitals rule to take the limit.
 
May 2010
274
67
Los Angeles, California
You need to be careful here. The residue at \(\displaystyle \zeta_k\) is

\(\displaystyle \lim_{z\rightarrow \zeta_k}\frac{(z-\zeta_k)f(z)}{z^n-a}\)

which is not what you have. (I'm also assuming \(\displaystyle f(z)\) is not 0 at these roots). If you want to apply L'Hopital's rule, you need to differentiate \(\displaystyle f(z)\) as well. However, these terms disappear after taking a limit. I don't think the exponents you have in your original answer look right.

\(\displaystyle \lim_{z\rightarrow \zeta_k}\frac{(z-\zeta_k)f(z)}{z^n-a}=\lim_{z\rightarrow \zeta_k}\frac{f(z)+(z-\zeta_k)f^\prime (z)}{nz^{n-1}}=\frac{f(\zeta_k)}{n\zeta_k^{n-1}}=\frac{f(\zeta_k)\zeta_k}{na}\).

Summing over this gives a slightly different answer to what you've posted (no \(\displaystyle n-1\) term in the exponent).
 
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May 2010
43
1
Actually, the step you took in evaluating the limit was accounted for in my expression of the residues, so I think we're in agreement there. Let's start with the expression \(\displaystyle 2\pi i\sum_{k=0}^{n-1}f(\zeta _k)/(n\zeta _k^{n-1})\), which is your expression for the residues.. I've just added the summation over them.

Substituting \(\displaystyle \zeta _k=a^{1/n}exp^{2\pi ik/n}\) gives

\(\displaystyle 2\pi i\sum_{k=0}^{n-1}\frac{f(\zeta _k)}{na^{(n-1)/n}}exp^{-2\pi ik(n-1)/n} \)

Which is essentially what I had originally posted (except for the "-" sign in the phase term)

So, I think you and I end up in the same place.. right?
 
May 2010
274
67
Los Angeles, California
OK, I see what you did. I think my form is a little simpler but both seem correct. Notice that \(\displaystyle \sum_{k=0}^{n-1}\zeta_k = 0\) and so if \(\displaystyle f(z)\) is constant, the contour integral will be 0.