Iranian Mathematical Olympiad

a.
We have
$$P^*(z) = \bar{a}_0 z^d + \bar{a}_1 z^{d-1} + \cdots + \bar{a}_d = z^d \left( \bar{a}_0 + \bar{a}_1 \frac{1}{z} + \cdots + \bar{a}_d \left( \frac{1}{z} \right)^d \right)$$
so
$$P^*(z) = z^d \overline{\left( a_0 + a_1 \frac{1}{z} + \cdots + a_d \left( \frac{1}{z} \right)^d \right)} = z^d \overline{P\left( \frac{1}{z} \right)}$$

b.
Let $z_1, z_2, \dots, z_n$ be roots (not necessarily distinct) of $q$. So
$$q(z) = (z - z_1) \cdots (z - z_n)$$
By assumptions of the problem for every $1 \le i \le n$ we know that $|z_i| \le 1$. Using the first part of the problem we have
$$q^*(z) = z^n \overline{q\left(\frac{1}{z}\right)} = z^n \overline{\left(\frac{1}{z} - z_1\right) \cdots \left(\frac{1}{z} - z_n\right)}$$
so we have
$$q^*(z) = (1 - z\overline{z_1}) \cdots (1 - z\overline{z_n})$$
Now assume that $r$ is a root of polynomial $Q$. Then
$$Q(r) = r^m q(r) + q^*(r) = 0$$
so
$$r^m(r - z_1) \cdots (r - z_n) = -(1 - r\bar{z}_1) \cdots (1 - r\bar{z}_n)$$
which implies
$$|r|^m |r - z_1| \cdots |r - z_n| = |1 - r\bar{z}_1| \cdots |1 - r\bar{z}_n|$$

Case 1: $|r| > 1$

We claim that for every $1 \le i \le n$, we have $|r - z_i| \ge |1 - r\bar{z}_i|$. We have
$$|r - z_i|^2 = (r - z_i)(\overline{(r - z_i)}) = |r|^2 - r\bar{z}_i + \bar{r}z_i + |z_i|^2 \\ |1 - r\bar{z}_i|^2 = (1 - r\bar{z}_i)(\overline{(1 - r\bar{z}_i)}) = 1 - r\bar{z}_i - \bar{r}z_i + (|r||z_i|)^2$$
So it is enough to prove that
$$|r|^2 + |z_i|^2 \ge 1 + (|r||z_i|)^2$$
Or equivalently
$$(|r|^2 - 1)(1 - |z_i|^2) \ge 0$$
Which is true. So our claim is proved. So according to the inequality we must have
$$|r|^m |r - z_1| \cdots |r - z_n| > |r - z_1| \cdots |r - z_n| \ge |1 - r\bar{z}_1| \cdots |1 - r\bar{z}_n|$$
Which is a contradiction. So $|r| \le 1$.

Case 2: $|r| < 1$

Same as the previous part we would have $|r-z_i| \le |1-r\bar{z}_i|$ which also leads to a contradiction and implies $|r| \ge 1$. Thus we can conclude that $|r| = 1$.

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