One of the most interesting paragraphs I have ever read

[showman]

I don't see any mistake in his derivation, the only caveat is he is talking about apparent impedance, real part and imaginary part. The key here is he is assuming a damped oscillation. However, I do not see a mistake in his math and the real part of the impedance is affected by the circuits reactance. Remarkable, isn't it?

As far as perpetual motion, I have some thoughts on that but prefer not to share them here. It goes well beyond EE and into physics. I could make an argument either way with well known physics.

For one, if you don't see any mistakes, can you show me how you derive a = 1/sqrt(4L/(r^2K)-1) from the equations at the bottom of page 343 and top of 344.

[Nominal Animal]

The equations of the PDF pages 343 and 344 are
$$r - a s - \frac{a}{1 + a^2} k = 0 \tag{1a}\label{1a}$$
$$s - \frac{k}{1 + a^2} = 0 \tag{1b}\label{1b}$$
$$s = 2 \pi N L \tag{2a}\label{2a}$$
$$k = \frac{1}{2 \pi N K} \tag{2b}\label{2b}$$
and the claim is that the solution \$a\$ (\$0 \lt a \in \mathbb{R}\$) is
$$a = \frac{1}{\sqrt{\displaystyle \frac{4 L}{r^2 K} - 1}} \tag{3}\label{3}$$
and that it can be derived from the preceding equations.

I do not think so.

If we substitute equations \$\eqref{2a}\$ and \$\eqref{2b}\$ into equation \$\eqref{1b}\$, we get
$$2 \pi N L - \frac{1}{(2 \pi N K)(1 + a^2)} = 0$$
and since we can safely assume \$0 \lt a \in \mathbb{R}\$ and \$L N \ne 0\$,
$$a^2 = \frac{1}{(2 \pi N K)(2 \pi N L)} - 1$$
and
$$a = \sqrt{\frac{1}{(2 \pi N K)(2 \pi N L)} - 1} \tag{4}\label{4}$$

Solutions \$\eqref{3}\$ and \$\eqref{4}\$ are equivalent if and only if
$$r = \pm \sqrt{\frac{4 L}{K} - (4 \pi N L)^2}$$
i.e. \$r\$ is a constant that does not depend on \$a\$.

However, if we solve equation \$\eqref{1a}\$ for \$r\$ we get
$$r = a s + \frac{a k}{1 + a^2}$$
If we solve equation \$\eqref{2a}\$ for \$s\$ we get
$$s = \frac{k}{1 + a^2}$$
Substituting the latter into the former yields
$$r = \frac{a k}{1 + a^2} + \frac{a k}{1 + a^2} = \frac{2 a k}{1 + a^2}$$
and substituting \$\eqref{2b}\$ into this yields
$$r = \frac{2 a}{(2 \pi N K)(1 + a^2)} = \frac{a}{(\pi N K)(1 + a^2)}$$
which means \$r\$ must be a function of \$a\$ if \$N\$ and \$K\$ are constants with respect to \$a\$.  That means the two solutions cannot be equivalent, and thus the purported solution \$\eqref{3}\$ is incorrect.

[showman]

That means the two solutions cannot be equivalent, and thus the purported solution \$\eqref{3}\$ is incorrect.

Yes, either that or the solution is indeed correct, but the preceeding equations are not.