Show that $$ \int_{|z|=1}(z+1/z)^{2m+1}dz = 2\pi i {2m+1 \choose m} $$ , for any nonnegative integer m.
I can't solve this problem..
I tried to find singularities but failed.
$(z+1/z)$ is not familiar to me.
Is there anyone to help?
Asked
Active
Viewed 577 times
2
-
I edited my question. – jakeoung Jun 12 '13 at 03:30
-
$z = 0$ is the only singularity. I suggest applying Cauchy's formula: http://en.wikipedia.org/wiki/Cauchy's_integral_formula#Theorem – Suugaku Jun 12 '13 at 03:32
-
1Here is the technique. – Mhenni Benghorbal Jun 12 '13 at 03:40
-
Sorry, I just noticed that there were some typos in my answer. Now they are fixed. – Hu Zhengtang Jun 12 '13 at 17:27
2 Answers
1
Due to binomial theorem, $$(z+\frac{1}{z})^{2m+1}=\frac{(z^2+1)^{2m+1}}{z^{2m+1}}=\frac{\sum_{k=0}^{2m+1}{2m+1 \choose k}z^{2k}}{z^{2m+1}}=\sum_{k=0}^{2m+1}{2m+1 \choose k}z^{2k-2m-1}.\tag{1}$$ By Cauchy's integral formula or direct calculation with $z=e^{it}$ for $|z|=1$, $$\int_{|z|=1}z^ndz = \left\{\begin{array}{cc} 2\pi i& n={-1}\\0 &n\ne{-1} \end{array}\right..\tag{2}$$ The conclusion follows from $(1)$ and $(2)$.
Hu Zhengtang
- 3,757
1
Hint: for $n\in\mathbb Z$ and a positively oriented contour: $$ \oint_{|z|=1}z^ndz = \begin{cases} 2\pi i & n=-1\\ 0 & \text{otherwise} \end{cases} $$ Now, what happens when you expand the binomial?
Ben Grossmann
- 225,327