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Find $$\Pi^{7}_{k=1}(z-e^{-k\pi i})$$

Calculating by hands seems possible but it requires a lot of work. Is there any tricks? Periodicity helps?

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2 Answers 2

up vote 4 down vote accepted

As you noted, periodicity is very helpful. $e^{-k\pi i} = e^{\pi i} = -1$ for odd $k$ and $e^{-k\pi i} = e^{2\pi i} = 1$ for even $k$. Your expression then simplifies to $$(z+1)^4(z-1)^3 = (z^2 - 1)^3(z+1)$$

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Thanks! It should be the fastest and most beautiful solution. –  John Hass Oct 28 '12 at 15:45

Using Euler's formula, $e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1,$ so $(e^{i\pi})^{2r+1}=-1$ and $(e^{i\pi})^{2r}=1$

$1\le k\le 2n+1 $ contains $n+1$ odd ,$n$ even values of $k$.

So, $$\Pi^{2n+1}_{k=1}(z-e^{-k\pi i})=\{z-(-1)\}^{n+1}(z-1)^n=(z+1)(z^2-1)^n$$

Here $n=3$.

$1\le k\le 2n $ contains $n$ odd ,$n$ even values of $k$.

So, $$\Pi^{2n}_{k=1}(z-e^{-k\pi i})=\{z-(-1)\}^n(z-1)^n=(z^2-1)^n$$

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