Derivation of Characteristic Impedance?
This seems the simplest mathematical way to derive characteristic impedance. Consider a "lump" of transmission line connected to the continuation of that transmission line (\$Z_0\$): -




R is series resistance of cable for a given length
L is series inductance of cable for a given length
G is parallel conductance of cable for a given length
C is parallel capacitance of cable for a given length
\$Z_0\$ to the right is the continuation of the cable


Therefore the impedance looking into the left is: -

$$Z_0 = R + j\omega L + Z_0||\dfrac{1}{G + j\omega C}$$

$$= R + j\omega L + \dfrac{\frac{Z_0}{G+j\omega C}}{Z_0 + \frac{1}{G+j\omega C}}$$ 

$$= R + j\omega L + \dfrac{Z_0}{1 + Z_0(G+j\omega C)}$$

$$Z_0[1 + Z_0(G+j\omega C)] = [R+j\omega L][1 + Z_0(G+j\omega C)]+Z_0$$

$$Z_0 + Z_0^2(G+j\omega C) = R+j\omega L + Z_0[(R+j\omega L)(G+j\omega C)]+Z_0$$

$$Z_0^2(G+j\omega C) = R+j\omega L + Z_0[(R+j\omega L)(G+j\omega C)]$$

The important thing next is to recognize that \$(R+j\omega L)(G+j\omega C)\$ is insignificant as the "lump" approaches zero length and we are left with: -

$$Z_0^2(G+j\omega C) = R+j\omega L$$ 

hence $$Z_0 = \sqrt{\dfrac{R+j\omega L}{G+j\omega C}}$$