Phase Contribution of Zeros in Z-plane
You're right that the angle $\Omega(\omega)$ in your sketch will equal $\pi$ for $\omega=\pi$. However, $\Omega(\omega)$ is not the phase angle of the system. In order to see this, let's write down the frequency response of a system with a single zero at $z_0=re^{j\theta}$:

$$H(e^{j\omega})=1-re^{j\theta}e^{-j\omega}=\frac{e^{j\omega}-re^{j\theta}}{e^{j\omega}}\tag{1}$$

From $(1)$, the phase angle is given by

$$\angle\{H(e^{j\omega})\}=\angle\{e^{j\omega}-re^{j\theta}\}-\angle\{e^{j\omega}\}=\Omega-\omega\tag{2}$$

So you indeed need the angle $\Omega$ as defined in your figures, but you also need to subtract the angular frequency $\omega$.

Let's assume $\theta=\pi$, as in your example. For $\omega=\pi$ (and $r&lt;1$) you get $\Omega=\pi$, and, consequently, a phase of $\Omega-\omega=0$. Note that if $r&gt;1$, you'll get $\Omega=0$ at $\omega=\pi$, and, consequently, a phase angle of $(-)\pi$.