What is the reason behind complex conjugate pairs in Linear Phase FIR filter analysis from the Pole Zero plot
The impulse response of a linear phase FIR filter is odd or even symmetrical. This has as a consequence that if $z_0$ is a zero, $1/z_0^*$ must also be a zero. If in addition to the linear phase property you also have real-valued coefficients, complex zeros always occur in complex conjugate pairs. So if a real-valued linear phase filter has a complex root $z_0$, you automatically get $z_0^*$, $1/z_0$, and $1/z_0^*$ as roots.
Complex roots on the unit circle just occur in complex conjugate pairs (because $z_0=1/z_0^*$, and real-valued roots just occur in pairs $z_0$ and $1/z_0$, unless they're on the circle (i.e., $z_0=1$ or $z_0=-1$).
In sum, for real-valued linear phase FIR filters, there are four types of &quot;clusterings&quot; of roots:

complex roots with $|z_0|\neq 1$ occur in groups of four: $z_0$, $z_0^*$, $1/z_0$, $1/z_0^*$
complex roots on the unit circle occur as complex conjugate pairs $z_0$ and $z_0^*$
real roots with $|z_0|\neq 1$ occur in pairs as $z_0$ and $1/z_0$
real roots on the unit circle ($z_0=1$ or $z_0=-1$ ) don't imply another root


Mathematically, for a real-valued length $N$ FIR filter the linear phase condition in the frequency domain is
$$H(z)=\pm z^{-(N-1)}H\left(\frac{1}{z}\right)\tag{1}$$
where $N$ is the filter length (number of taps). From $(1)$ it is clear that if $z_0$ is a zero of $H(z)$ then also $1/z_0$ must be a zero.