prove that a process is an MA(q)

Question

We are given the following polynomial : $$\Theta(z) = 1 + \theta_1 z + ...+ \theta_q z^q $$ and we suppose that $(z_t)_{t \in \mathbb{Z}}$ is weakly stationnary and its spectral measure is given by : $$ \mu_z(d\omega) = |\Theta(e^{i\omega})|^2\sigma^2/2\pi d\omega$$ and that $\Theta$ has no root in the unit circle. We want to prove that $(z_t)$ is an $MA(q)$.
I used the inverse filtering theorem to show that there exists $(\epsilon_t)_t$ weakly stationnary such that $z_t = \Theta(L) \epsilon_t $ where L is the backward shift operator. Indeed $\sum_{-\infty}^{\infty} |\theta_i| \lt \infty $ and $\int_{-\pi}^{\pi}\frac{1}{|\Theta(e^{-i\omega})|^2}\mu_z(d\omega) < \infty$. Furthermore, $\gamma_{\epsilon}(h) = \sigma^2 \mathbb{1}_{h=0}$ but i have yet to prove that the expectation of $(\epsilon_t)$ is null, and i have no clue on how to do it.
Also, i don't see where we use the hypothesis that $\Theta$ has no root in the unit circle since $|\Theta(e^{-i\omega})|^2$ is contained in the spectral measure of $z_t$ and it simplifies inside the integral which makes it converge. So it seems to me that this hypothesis is not necessary. Thank you for your help