Let $P(z)=\sum_{k\in \mathbb{Z}}p_{k}z^{k}$ and define $Q(z)=-z\overline{p(-z)}$. for $\left | z \right |=1$, show that $Q(z)=1/2\sum_{k\in \mathbb{Z}}(-1)^k\overline{p}_{1-k}z^{k}$.
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Please? And what are the $;p_k$'s ? And what have you done so far to tackle this problem? – Timbuc Jan 22 '15 at 12:56
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For some finite number of constants $p_{k}$ – Ehsan zarei Jan 22 '15 at 13:19
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....and? What did you try? For example, you have that $;1=|z|=|z|^2=z,\overline z\iff \overline z=\frac1z;$ and etc. – Timbuc Jan 22 '15 at 13:26
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Observe that
$$|z|=1\iff 1=|z|^2=z\overline z\iff z=\frac1{\overline z}$$
$$Q(z)=-z\overline{p(-z)}=-z\overline{\sum_{k\in\Bbb Z}p_k(-z)^k}=-z\sum_{k\in\Bbb Z}\overline{p_k}\overline{(-z)^k}=-\overline{z^{-1}}\sum_{k\in\Bbb Z}\overline{p_k}\overline{(-z)^k}=$$
$$=\sum_{k\in\Bbb Z}\overline{p_k}\overline{(-z)^{k-1}}=\sum_{k\in\Bbb Z}(-1)^{k-1}\,\overline{p_k}\,\overline{z^{k-1}}$$
Now finish the problem.
Timbuc
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Thanks Timbuc but $Q(z)=1/2\sum_{k\in \mathbb{Z}}(-1)^k\overline{p}_{1-k}z^{k}$ and Your result is different ? – Ehsan zarei Jan 22 '15 at 13:43
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@ehsanzarei There is no "result" in my answer: you still have some work to do (very little, btw.). I showed you my way, now you complete. – Timbuc Jan 22 '15 at 14:05