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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?

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    @J$a$ck: depending on how you compute the derivatives, that will either end up $b$eing a more complicated version of e$x$p$a$nding it out or a more complicated version of applying the $b$inomial theorem.2011-05-25

2 Answers 2

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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.

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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.