Reg Edwards wrote:
Write it down. It will then be obvious, to make the angle of Zo equal to
zero it is necessary only that the angle of R+j*Omega*L be made equal to the
angle of G+j*Omega*C.

Reg, there is no contradiction between what I have said and what you have said.

Continuing the discussion - For average transmission lines used on HF
frequencies, "... the value of G ... is likely to be too small to affect the
attenuation factor ..." Quote from "Transmission Lines" by Chipman, page 94.

Some of Chipman's calculations indicate that, for a typical 10 MHz example,
R is about 0.1 ohm/meter while G is about 0.9 micromhos/meter. That's about
a 100,000:1 ratio making G negligible as far as attenuation factor goes.

The attenuation factor depends almost entirely on R, the series resistance
parameter. G, the parallel conductance parameter, has a negligible effect
on the attenuation factor at HF.

Since, at HF, the attenuation factor consists almost entirely of series
resistance, and since the attenuation factor is identical for voltage and
current, it logically follows that the series resistance is primarily
responsible for the attenuation of the current.

Or even more simple, for Zo to be purely resistive, G = C*R/L

Actually, that is only an approximation for low-loss lines.
--
73, Cecil, W5DXP