IIR Hilbert Transformer I'm beginning to explore discrete Hilbert transformers - ways to achieve 90°. phase shift across a band of perhaps 6 kHz at a 44.1 kHz sampling rate. I'm trying to stick with IIR filters in order to save computation time - in addition to the fact that I only need to phase shift just a band of frequencies.

I saw this post where the author Ross Wilkinson alludes to creating an IIR based Hilbert transformer by setting the transfer function's zeros to: 

\begin{equation}
{(q_n)}^{k}_{n=0},\quad ‎‎q_n = e^{\frac{\pi}{2^n}}\quad
\end{equation}

The RHS poles inside the unit circle (0 &lt; x &lt; 1) are the inverse of the zeros:
\begin{equation}
{(p_n)}^{k}_{n=0},\quad p_n = \frac{1}{q_n}
\end{equation}

It would be great if somebody could point me towards another resource that would help explain where that comes from in greater detail.

FYI, I'm trying to make a single side band modulator for audio frequencies. It should be able to process in real time on a laptop, with a sampling rate of 44.1 kHz.
This is achievable with two parallel all pass filters.
The two all pass filters synthesize an odd ordered low pass filter whose pass band extends from -90º to +90º in the z-domain. (I will discuss this below).
$$
G_{lowpass}(z) = \frac{A_0(z)+z^{-1}A_1(z)}{2}
$$
The low pass filter is then rotated by +90º so that its pass band extends from 0º to 180º, which approximates the Hilbert transform. Rotation mathematically is:
$$
H_{Hilbert}(z) =G_{lowpass}(-jz)
$$
As a consequence, one of the all pass filters becomes completely imaginary - this is the Hilbert transformed signal path. By necessity, since the filter is approximating this Hilbert transform, then the imaginary branch must be applying a 90º phase shift relative to the output of $A_0(z)$.
$$
H_{Hillbert}(z)=G_{lowpass}(-jz)=\frac{A_0(z)+jz^{-1}A_1(z)}{2}
$$

Example with Elliptical Filter
The following example was taken from Handbook for Digital Signal Processing pg. 920
You can also use this program that I interpreted to design a new half-band elliptic filter with a higher order.
Starting with a real, half-band elliptical low pass filter $G(z)$ whose frequency response satisfies
$$
1-\delta_1\leq|G(e^{j\omega})|\leq1\&gt;\&gt;\&gt;\&gt;\&gt;for\&gt;\&gt;0\leq\omega\leq\omega_p
$$
$$
\&gt;\&gt;\&gt;|G(e^{j\omega})|\leq\delta_2\&gt;\&gt;\&gt;\&gt;\&gt;for\&gt;\&gt;\omega_s\leq\omega\leq\pi
$$
To obtain half band we choose:
$$
\omega_s+\omega_p=\pi
$$
$$
(1-\delta_1)^{2}+\delta_2^{2}=1
$$
But I just used the examples given seventh order poles or an elliptical low pass half-band filter:
$$
z=0,\&gt;\&gt;\&gt;z=\pm j0.436688,\&gt;\&gt;\&gt;z=\pm j0.743707,\&gt;\&gt;\&gt;z=\pm j0.927758
$$
$$
G_{lowpass}(z) = \frac{A_0(z)+z^{-1}A_1(z)}{2}
$$
Distribute the poles between the two all pass filters like so:

$$
A_0(z)=\frac{z^{-2}+0.190696}{1+0.190696z^{-2}}\cdot\frac{z^{-2}+0.860735}{1+0.860735z^{-2}}
$$
$$
A_1(z)=\frac{z^{-2}+0.553100}{1+0.553100^{-2}}
$$
Rotate by 90º
$$
H_{Hilbert}(z)=G_{lowpass}(-jz)=\frac{A_0(-jz)+jz^{-1}A_1(-jz)}{2}
$$
$$
A_0(-jz)=\frac{0.190696-z^{-2}}{1-0.190696z^{-2}}\cdot\frac{0.860735-z^{-2}}{1-0.860735z^{-2}}
$$
$$
A_1(z)=\frac{0.553100-z^{-2}}{1-0.553100^{-2}}
$$
Phases of $A_0(z)$ and $A_1(z)$:

Magnitude Response of $H_{Hilbert}(z)$:

Phase Difference Between $A_0(z)$ and $A_1(z)$ Branches: