From your more recent posting, I misinterpreted what you mean by
"driving point impedance". It's simply the impedance of the transmission
line, not the impedance from the load looking back toward the source,
nor the source impedance. Based on that interpretation, I'll try again.

Peter O. Brackett wrote:
. . .
Now at any particular frequency w = 2*p*f you will find that this general
complex
surge impedance Zo evaluates to a complex number, say Zo(jw) = r + jx.
Later
let's let r = 50 Ohms and x = 5 Ohms so that we can work out a numerical
example.

Consider either a semi-infinite length of this Zo line, or even a finite
length of the Zo line
terminated in an impedance equal to Z0. I am sure that you will agree that
both the
semi-infinite Zo line or the finite length Zo line terminated in Zo have the
same
driving point impedance namely Zo.

That is the impedance of the line. Ok.

Now excite this semi-infinte Zo line by an ideal generator of open circuit
voltage
Vi = 2*a behind an impedance equal to the surge impedance Zo. In other
words
this is a Thevenin generator of ideal constant voltage 2a behind a complex
impedance
of Zo.

Oops. No it's not. Consider circuit "A", a voltage source in series with
a lumped impedance equal to r + jx. And circuit "B", a voltage source in
series with a transmission line of length, say, one wavelength, with
characteristic impedance Z0 = r + jx. The two are not at all equivalent.
To show that they're not, connect a 50 ohm resistor to the output of
each. The resistor current in circuit "A" is Vi/(50 + r + jx). The
current in circuit "B" is Vi/50.

David Robbins recently made several postings explaining that these two
circuits are not equivalent. Hopefully the above example illustrates that.

Can't you do your "proof" with, say, a one wavelength transmission line
of characteristic impedance Z0?

Roy Lewallen, W7EL