1

How can I show the following, for $n\geq 0$:

$$ \frac{1}{2\pi} \oint_{\ \Gamma} \frac{1}{z} \left(z + \frac{1}{z}\right)^{2n} dz $$

using a contour $\Gamma$ defined as the unit circle centered at the origin and oriented counterclocwkise.

Ref. Complex Analysis by M.W. Wong

user14685
  • 121
  • You're missing an $i$ in the denominator of the factor before the first integral. Do you already know the residue theorem? – Daniel Fischer Apr 27 '14 at 13:19
  • ... so I should be doing a Laurent series expansion to get the residues at the poles and if my memory is good, the integral is equal to the sum of the residues? – user14685 Apr 27 '14 at 13:26
  • Yes (if you fix the missing $i$). If you look at the integrand a little more closely, you will see that the poles are few in number, and the residue easily computed. – Daniel Fischer Apr 27 '14 at 13:29
  • Wong's book explicitely states $2\pi$... perhaps its is to cancel out with the $2\pi i$ of the residue formula... – user14685 Apr 27 '14 at 14:07
  • 1
    The first integral. That ought to be $$\frac{1}{2\pi i}\int_\Gamma \frac{1}{z}\left(z+\frac{1}{z}\right)^{2n},dz.$$ – Daniel Fischer Apr 27 '14 at 14:09

1 Answers1

0

The residue theorem asks for the $z^{-1}$ term of $z^{-1}(z+z^{-1})^{2n}$, which is the same as the constant coefficient of $(z+z^{-1})^{\color{Red}{2n}}$. If I said we get $\binom{\color{Red}{2n}}{n}$ would you know how the binomial appears?

anon
  • 151,657