0
$\begingroup$

Consider the following question:

What's the coefficient of $z^4$ where

$$\Pi_{k=1}^{7}(z-e^{ik\pi}), \quad z\in{\bf C}$$

I am wondering whether there is some "trick" for solving the problem above quickly, instead of simply expanding the product or using the Taylor expansion. More generally,

What's the coefficient of $z^m$ where

$$\Pi_{k=1}^{n}(z-e^{ik\pi}), \quad z\in{\bf C}$$

What's the topic to which the above question is related in complex analysis and where such formula might appear?

  • 4
    This isn't a problem in complex analysis. Do you know what $e^{i \pi}$ is?2011-05-25
  • 0
    @Qiaochu: Oops, I found I raised a stupid question here. I think your nice comment is enough to be the answer. If one replace $e^{ik\pi}$ to be $z_k$, which was at first in my mind, things might be not that obvious.2011-05-25
  • 2
    @Jack: then no. The coefficient of $z^4$ is nothing more and nothing less than a certain polynomial in the $z_k$. When $z_k$ takes a special form there are various options depending on the special form.2011-05-25
  • 1
    @Qiaochu: I'm curious about your edited comment---why this is not a problem in complex analysis? And this has confused me a long time. In the undergraduate courses in some countries, complex analysis is sometimes called "Theory of complex variable functions". Some students might relate the topic about complex numbers to complex analysis.2011-05-25
  • 2
    @Jack: beyond knowing what $e^{i \pi}$ is, this is a completely elementary problem, and the intent of the problem is not (I assume) to test your knowledge of the value of $e^{i \pi}$. The fact that $z$ is stipulated to lie in $\mathbb{C}$ is irrelevant.2011-05-25
  • 0
    @Qiaochu: Hmm, binomial coefficients is enough. What I thought is that one can use the Taylor expansion at $z=0$ to calculate the coefficients.2011-05-25
  • 0
    @Jack: depending on how you compute the derivatives, that will either end up being a more complicated version of expanding it out or a more complicated version of applying the binomial theorem.2011-05-25

2 Answers 2

5

As mentioned above, $e^{ik} \in \{-1,1\}$. Therefore, your product is $(z^2-1)^{n/2}$ when $n$ is even and $(z^2-1)^{(n-1)/2}(z-1)$ if $n$ is odd. Expanding and using binomial coefficients you can get a closed form for the result.

2

You can use contour integration and Cauchy's formula to solve the problem as follows:

If we integrate around a contour, we pick up information about the residues of our function. Therefore, if we want to convert information about $z^4$ into information contained in residues, we divide by $z^5$. Then, by integrating over over any circle containing $0$, we obtain the desired result.

Of course, this is probably slower than just using the binomial theorem.