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When I look at the Pole/Zero diagrams in the DSP books it appears that whatever pole or zero is above the real x-axis from:

$$\omega=0 \rightarrow \pi$$

is always reflected as a mirror image below the x-axis from:

$$\omega=0 \rightarrow -\pi$$.

(And whatever pole or zero is on the real axis with $Im\{H(z)\}=0$ is not reflected anywhere.)

For instance, if you have a zero at $e^{j\omega_0}$ and ($\omega_0\ \ne \pi$, and $\omega_0\ \ne 0$) then you are always going to get a zero that is below the x-axis at $e^{-j\omega_0}$ regardless what type of filter or function the system function H(z) is implementing.

I'm wondering what law or rule this relationship comes from? and are there any cases where this is not true for a realizable system function of H(z)?

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No it does not (always posses symmetry).

However, for real systems whose coefficients $a_k ~,~ b_k$ of system function,

$$H(z) = \sum_{n} h[n] z^{-n} = \frac{\sum_{k=0}^{M}b_k z^{-k}}{\sum_{k=0}^{N} a_k z^{-k}} $$ are real, then accordingly its roots (which are the poles and zeros of $H(z)$) are either real or complex-conjugates. And because of this conjugation, the poles (or zeros) on the Z-plane posses the vertical symmetry as you've recognised.

On the contrary, for complex systems whose system function $H(z)$ is a complex polynomial of complex coefficients, we cannot constraint the roots of denominator and numerator of $H(z)$ to be real or complex-conjugate, and hence they can posses non-symmetric poles and zeros...

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  • $\begingroup$ So basically if h[n] is real, then you will always get either real or complex conjugates poles and zeros? but if h[n] is complex then its not required to have complex conjugates for poles and zeros. Could I say that also? $\endgroup$ – MrCasuality Jan 15 at 18:34
  • $\begingroup$ yes, exactly... $\endgroup$ – Fat32 Jan 15 at 18:44

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