show that $Q(z)=1/2\sum_{k\in \mathbb{Z}}(-1)^k\overline{p}_{1-k}z^{k}$

Question

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 $Q(z)=1/2\sum_{k\in \mathbb{Z}}(-1)^k\overline{p}_{1-k}z^{k}$.

Answer 1

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.