Magnitude response of mirrored (with respect to unit circle) poles and zeros I just want to check that my understanding about the following paragraph from Optical Filter Design
and Analysis by Christi K. Madsen, Jian H. Zhao is correct:


A filter’s magnitude response is equal to the modulus of its transfer function,
$|H(z)|$, evaluated at $z = e^{j\omega}$. Based on the pole/zero representation of $H(z)$, only the
distance of each pole and zero from the unit circle, i.e. $|e^{j\omega}-z_m|$ or $|e^{j\omega}-p_n|$, affects the magnitude response. Consequently, a zero that is located at the mirror image
position about the unit circle, i.e. $1/z_m^*$, cannot be differentiated from $z_m$ based on
the magnitude response.


Let us assume a filter with just one zero $c_1=|c_1|e^{j\phi}$ so that its transfer function is:
$$
H(z)=1-c_1 z^{-1}=1-\lvert c_1\rvert e^{j\phi} z^{-1}
$$
Its mirrored zero is $c_2=1/c_1^*=\lvert c_1\rvert^{-1}e^{j\phi}$ and its transfer function is
$$
H_m(z)=1-c_2 z^{-1}=1-\lvert c_1\rvert^{-1}e^{j\phi} z^{-1}
$$
Evaluating the transfer function at $z=e^{j\omega}$ and computing $H(e^{j\omega})H^*(e^{j\omega})$ we get the power response:
$$
\big\lvert H(e^{j\omega)}\big\rvert^2=\big\lvert 1-\lvert c_1\rvert e^{j(\phi-\omega)}\big\rvert^2=1 +\lvert c_1\rvert^2-2\lvert c_1\rvert\cos(\phi-\omega)
$$
and
$$\big\lvert H_m(e^{j\omega})\big\rvert^2=\bigg\lvert 1-\frac{1}{\lvert c_1\rvert}e^{j(\phi-\omega)}\bigg\rvert^2=1 +\frac{1}{\lvert c_1\rvert^2}-2\frac{1}{\lvert c_1\rvert}\cos(\phi-\omega)
$$
$\big\lvert H_m(e^{j\omega})\big\rvert^2$ can also be expressed as:
$$
\big\lvert H_m(e^{j\omega})\big\rvert^2=\frac{1}{\lvert c_1\rvert^2}\Big[1 +\lvert c_1\rvert^2-2\lvert c_1\rvert\cos(\phi-\omega)\Big]=\frac{1}{\lvert c_1\rvert^2}\lvert H(e^{j\omega)}\rvert^2
$$
which is just the power response of the first filter scaled by a factor of $\lvert c_1\rvert^{-2}$. Is this analysis right?
You're right, the contribution of a zero $z_0=re^{j\phi}$ to the squared magnitude response is
$$\big|e^{j\omega}-re^{j\phi}\big|^2=1-2r\cos(\omega-\phi)+r^2\tag{1}$$
From $(1)$ it is clear that another zero resulting in the same frequency dependence as in $(1)$ must have the same phase angle $\phi$. Assuming $z_1=Re^{j\phi}$ we get
$$\big|e^{j\omega}-Re^{j\phi}\big|^2=1-2R\cos(\omega-\phi)+R^2=R^2\left(1-\frac{2}{R}\cos(\omega-\phi)+\frac{1}{R^2}\right)\tag{2}$$
Comparing $(1)$ and $(2)$ we get $R=1/r$ and a squared magnitude scaling of $R^2=1/r^2$. Consequently, the other zero $z_1$ is given by
$$z_1=\frac{e^{j\phi}}{r}=\frac{1}{z_0^*}\tag{3}$$
and
$$\big|e^{j\omega}-z_0\big|=r\left|e^{j\omega}-\frac{1}{z_0^*}\right|\tag{4}$$