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I need to use the following identities for poisson integral but i can't guz i don't know how to prove them.



I would appreciate any help

what i did :

Let $\alpha >1\,\quad \alpha^{2n} =1$

\begin{align*} \alpha^{2n} &=1\\ (\rho.e^{i\phi})^{2n} &=1.e^{i.0} \text{ with $\rho \in \mathbb{R}^{+}_{*},\ \phi \in \mathbb{R}$}\\ \rho^{2n}.e^{i.2n.\phi}&=1.e^{i.0}\\ \rho^{2n}&=1, \quad e^{i.2n\phi}=e^{i.0}\\ \rho&=1, \quad 2n.\phi=0[2\pi]\\ \rho&=1, \quad \exists k\in \mathbb{Z} \ 2n.\phi=2.k.\pi\\ \rho&=1, \quad \phi=\frac{2.k.\pi}{2n} \text{ Then } \alpha_{k}=e^{i\frac{2.k.\pi}{2n}}\\ \text{Let } k &= t + 2nq \text{ or } t\in \{0,1,\ldots,2n-1\} \text{ and } q \in \mathbb{Z}.\\ &\text{ Then } \alpha_{k}=e^{i\dfrac{2.t.\pi}{2n}+2.\pi.q}=Z_{t} \\ S&=\{Z_{k}=e^{i\dfrac{2.k.\pi}{2n}}|k\in \{0,1,\ldots,2n-1\} \} \end{align*}

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up vote 2 down vote accepted

First Identity: Let $$p(x) = x^n + a_{n-1} x^{n-1}+ \ldots+a_0$$ a polynomial function of order $n$. Then, by the fundamental theorem of algebra, we have $$p(x) = \prod_{j=1}^n (x-x_j)$$ where $x_1,\ldots,x_n$ are the roots of $p$. Apply this to $p(x) := x^{2n}-1.$

Second Identity:

Write $$\prod_{k=0}^{2n-1} \left(\alpha - e^{\imath \frac{2k\pi}{2n}} \right) = \underbrace{\left(\prod_{k=0}^0 \dots \quad \cdot \prod_{k=n}^n \dots \right)}_{(\alpha-1)(\alpha+1)} \left( \prod_{k=1}^{n-1} \dots \quad \cdot \prod_{k=n+1}^{2n-1} \dots \right).$$

Now note that

$$\exp \left( 2\pi \imath \frac{2n-\ell}{2n} \right) = \exp \left( 2\pi \imath - 2\pi \imath \frac{\ell}{2n} \right) = \exp \left(- 2\pi \imath \frac{\ell}{2n} \right) \tag{1}$$

as $x \mapsto e^{\imath x}$ is periodic. For $k \in \{n+1,\ldots,2n-1\}$ we write $k = 2n-\ell$ where $\ell \in \{1,\ldots,n-1\}$. Applying $(1)$, we get

$$\prod_{k=n+1}^{2n-1} \left(\alpha - e^{\imath \frac{2k\pi}{2n}} \right) = \prod_{\ell=1}^{n-1} \left(\alpha - e^{- \imath \pi \frac{\ell}{n}} \right).$$

This finishes the proof.

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Fundamental Theorem of Algebra--not Calculus. – Jared Jun 7 '14 at 6:55
@Jared Of course; thanks! – saz Jun 7 '14 at 6:55
@Educ Let $k \in \{1,\ldots,n-1\}$. Then $$\exp \left(-\imath \frac{k\pi}{n} \right) = \exp \left(-\imath \frac{k\pi}{n} + \imath 2k\pi \right) = \exp \left(\imath \frac{(2k-1) \pi}{n}\right).$$ – saz Jun 7 '14 at 8:47
@Educ I have added the proof of the 2nd identity. – saz Jun 7 '14 at 12:17
@Educ Yes, but the multiplication is commutative, i.e. $a \cdot b = b \cdot a$. So $$\prod_{k=1}^n a_k = a_1 \cdot a_2 \cdots a_n = \prod_{k=n}^1 a_k.$$ – saz Jun 7 '14 at 14:01

The first line is a statement about roots of unity and factorization. The numbers $e^{2\pi i k / n}$ are precisely the $n$th roots of unity. (And the factor theorem, I suppose).

The second one is just a rearrangement of terms from the first. If you'd like an alternate view, you might think of $\alpha$ as a variable, and consider each side as polynomials in $\alpha$. Then since they have the same roots and the same value at $\alpha = 0$, they are the same polynomial.

In short: you should read up on roots of unity and possibly the factor theorem.

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