Phase response for conjugate zeros
If you have conjugate symmetric poles and zeros then the coefficients of the system (i.e., the filter coefficients, or, equivalently, the coefficients of the corresponding difference equation) are real-valued. E.g., if the zeros are $z_0$ and $z_0^*$ then the numerator polynomial of the transfer function $H(z)$ is (up to a scaling factor)

$$N(z)=(z-z_0)(z-z_0^*)=z^2-(z_0+z_0^*)z+|z_0|^2=z^2-2\text{Re}(z_0)z+|z_0|^2\tag{1}$$

Note that the coefficients in $(1)$ are all real-valued. The same is true for the denominator polynomial. So the frequency response (i.e., the transfer function evaluated at $z=e^{j\omega}$) can be written as

$$H(e^{j\omega})=\frac{b_0e^{2j\omega}+b_1e^{j\omega}+b_2}{e^{2j\omega}+a_1e^{j\omega}+a_2}\tag{2}$$

with all coefficients $a_i$ and $b_i$ being real-valued. Consequently, the following equation must hold:

$$H(e^{j\omega})=H^*(e^{-j\omega})\tag{3}$$

If you write $H(e^{j\omega})$ in polar form $H(e^{j\omega})$=$M(\omega)e^{j\phi(\omega)}$ with magnitude $M(\omega)$ and phase $\phi(\omega)$ you get

$$M(\omega)e^{j\phi(\omega)}=M(-\omega)e^{-j\phi(-\omega)}\tag{4}$$

From $(4)$ it follows that for conjugate symmetric poles and zeros the magnitude of the frequency response must be an even function of frequency, and the phase must be an odd function of frequency, i.e., $M(\omega)=M(-\omega)$ and $\phi(\omega)=-\phi(-\omega)$.