Show there exists $z \in \mathbb{U}$ such that $\prod_{k=1}^n (z - a_k) \in \mathbb{U}$ The exercise
Let $n\geq 3$ and $a_1, \dots, a_n \in \mathbb{U}$. Show there exists $z \in \mathbb{U}$ such that $\prod_{k=1}^n (z - a_k) \in \mathbb{U}$.
My try
Writing $z = e^{i \theta}$ and $a_k = e^{i \theta_k}$, I get that the module of the product is $2^n\prod_{k=1}^n | \sin(\frac{\theta - \theta_k}{2}) |$, using the fact that $e^{i \theta} - e^{i \theta_k} = 2 i \sin(\frac{\theta - \theta_k}{2}) e^{i \frac{\theta+\theta_k}{2}}$. I then need to find a value of $\theta$ that would make this number equal to $1$. Clearly, the function $f : \theta \mapsto 2^n \prod_{k=1}^n | \sin(\frac{\theta - \theta_k}{2}) |$ is continuous over $\mathbb{R}$. I would like to use the intermediate value theorem, but I have to show first $f$ takes at least one value $<1$ and another $> 1$.
Your help is welcome!
 A: $f(\theta_k) = 0$ for each $k$, so that it remains to show that $f(\theta) \ge 1$ for some $\theta$. Here helps the complex analysis:
$$
 g(z) = (z-a_1) \cdots (z-a_n)
$$
is holomorphic in $\Bbb C$. The maximum modulus principle implies that $|g|$ attains its maximal value on the closed unit disk at a point on the boundary:
$$
 1 = |g(0)| \le \max_{|z|=1} |g(z)| = |g(e^{i\theta})|
$$
for some $\theta \in [0, 2 \pi]$. (Actually the inequality is strict because $g$ is not constant, but that is not needed here.)
A: Without holomorphic functions, one can use the well-known fact (and easy to prove by the doubling formula, noting that the function is integrable by calculus considerations) that $\int_0^{\pi}\log(2|\sin \theta|)d \theta=0$.
Changing variables we get $\int_0^{2\pi}\log(2|\sin \theta/2|)d \theta=2\int_0^{\pi}\log(2|\sin t|)dt=0$ and $\log(2|\sin \theta/2|)$ is periodic with period $2\pi$, so this means $\int_0^{2\pi}\log(2|\sin (\theta-\theta_n)/2|)d \theta=0$ for any fixed $\theta_n$
But now if $f(\theta)=2^n\prod_{k=1}^n | \sin(\frac{\theta - \theta_k}{2}) | \ge 0$ integrable and periodic on $[0,2\pi]$, the above means $\int_0^{2\pi} \log f(\theta)d\theta=0$ and since $f(\theta)$ is not identically $1$ (as it is zero near $n$ points and continuous) we get that there is $\psi, \log f(\psi) >0$ hence $f(\psi)>1$, so by the intermediate value theorem ($f$ is zero at $\theta_k$) there is $\theta, f(\theta)=1$ and we are done!
