I certainly hope I managed to decrypt what's being asked here. If not, please clarify so I can provide a more relevant answer.
Let's start with the definition of the lossy and lossless models:
In lossy model the resistance of the line itself matters for the calculation! Basically we're running such voltage through the line that even if we just took replaced the signal with a DC voltage equal to the RMS voltage of our signal, we'd still have losses.
Using the lossless model, if we replaced the signal with DC voltage equal to RMS voltage of the signal, we'd have no considerable losses. In this case, only reactive components matter and the resistance of the line itself isn't important.
This brings us to the basic equation for the transmission line characteristic impedance:
\$Z_0=\sqrt{ \frac {R + j \omega L} {G + j \omega C}}\$
Keep in mind that G is conductance in parallel with the capacitance, and that R is resistance in series with the inductance. In the lossless model, we don't have the two of them, so their values are taken as zero. We can then rewrite the equation as
\$Z_0=\sqrt{ \frac {0 + j \omega L} {0 + j \omega C}}\$
which is equivalent to
\$Z_0=\sqrt{ \frac {j \omega L} {j \omega C}}\$
Which we can rewrite as
\$Z_0=\sqrt{ \frac {j \omega} {j \omega } \frac {L}{C}}\$
The \$j \omega\$ parts prodce one and we get in the end only
\$Z_0=\sqrt{\frac {L}{C}}\$
So the lossless model doesn't have the imaginary part because the imaginary units cancelled themselves out.
Impedance matching is tricky, but the role of a quarter wave transmission line is to map from one impedance to another. The actual impedance of the line will not match either the input or the output impedance - this is entirely expected.
However at a given frequency, when a correctly designed quarter wave line is inserted with the correct impedance, the output impedance will appear to the input as perfectly matched. In your case, the transformer will make the \$20\Omega\$ impedance appear as if it is a \$100\Omega\$ impedance meaning no mismatch. Essentially it guides the waves from one characteristic impedance to another.
The easiest way to visualise this is on a Smith chart, plot the two points 0.4 (\$20\Omega\$) and 2 (\$100\Omega\$). Then draw a circle centred on the resistive/real axis (line down the middle) which intersects both points. You will find that this point is located at 0.894 (\$44.7\Omega\$) if your calculations are correct. This is shown below at \$500\mathrm{MHz}\$, but the frequency is only important when converting the electrical length to a physical length.
What a quarter wave transformer does is rotate a given point by \$180^\circ\$ around its characteristic impedance on the Smith chart (that's \$\lambda/4 = 90^\circ\$ forward plus \$90^\circ\$ reverse).
Exactly why it does this is complex. But the end result of a long derivation is that for a transmission line of impedance \$Z_0\$ connected to a load of impedance \$Z_L\$ and with a length \$l\$, then the impedance at the input is given by:
$$Z_{in}=Z_0\frac{Z_L+jZ_0\tan\left(\beta l\right)}{Z_0+jZ_L\tan\left(\beta l\right)}$$
That is an ugly equation, but it just so happens if the electrical length \$\beta l\$ is \$\lambda/4\$ (\$90^\circ\$), the \$\tan\$ part goes to infinity which allows the equation to be simplified to:
$$Z_{in}=Z_0\frac{Z_0}{Z_L}=\frac{(Z_0)^2}{Z_L}\rightarrow Z_0=\sqrt{\left(Z_{in}Z_L\right)}$$
Which is where your calculation comes from.
With the quarter wave transformer in place, the load appears as matched to the source. In other words, the transformer matches both of its interfaces, not just the input end.
You can also see from this equation why the transformer only works for a single frequency - because it relies on the physical length being \$\lambda/4\$. You can actually (generally using advanced design tools) achieve an approximate match over a range of frequencies - basically a close enough but not exact match.
Best Answer
Phil, the real part of the propagation constant is the attenuation constant and this equals: -
\$Re\sqrt{(R+jwL)(G+jwC)}\$ and not the formula you have in your question.
The formula you have used is for characteristic impedance.
This wiki page should confirm this (right at the bottom): -
So, if you do the math at low frequencies (to make life easier) you see that the attenuation constant becomes \$\sqrt{RG}\$ and if R=G=1 then you have a constant of 1 and a lousy highly lossy line. A lossless line has a re(propagation constant) of zero.