@orolo should \$ z\cdot e^{i\theta}\$ be \$ \left | z \right |\cdot e^{i\theta} = \left | z \right |\angle\theta \$, where \$ \left | z \right |= \left | a+i*b \right |=\sqrt{ a^{2}+b^{2} } \$ or am I now mixing something.
It doesn't matter, the cable impedance z can be any impedance (except 0), and the load is a phasor with the same module and forming an angle theta with z. If the cable impedance is real, as in Simon's problem, things can be written as above, but for calculation purposes it works the same with arbitrary z. Thinking of Smith charts, the unit circle is being mapped into the imaginary axis; only the angle between impedances matters.
@Simon
To compute the impedance of the line + load, two things matter:
1) The length of the cable causes a delay from source to load, which appears as factor B in your equation.
2) Part of the signal may be reflected back to the source. Since the reflection coefficent is (ZL - Z0)/(ZL + Z0), it will only be zero (no reflection) if ZL = Z0. So there will be reflections if ZL and Z0 are different
as complex numbers. So it is not enough that Z0 and ZL have the same module, they must also have the same argument.
Imagine you connect a sinusoidal function generator to the cable, giving a voltage Vi. To compute the impedance, you must obtain Vi/Ii, the ratio of voltage to current into the cable. If you leave the generator working long enough to reach a steady state, you get:
a. The generator injects a signal Vi that propagates from source to load along the cable. The current at the source generated by this signal is Vi/Z0.
b. When the injected signal reaches the load, it will have been delayed by an angle B, since it travelled along the line: let's call it Vi(B). If Z0 and ZL are mismatched, a reflected wave will arise, with value G*Vi(B), where G is the reflection coefficient. That reflected wave will travel back to the source, getting an additional delay B when it reaches the source.
c. So back at the source, you have a reflected signal G*Vi(2B), which causes a current G*Vi(2B)/Z0 that must be added to the injected current Vi/Z0.
d. The total current at the source is: Ii = Vi/Z0 + G*Vi(2B)/Z0. Injected current plus reflected current, doubly delayed.
From there, using Zi = Vi/Ii, G = (ZL - Z0)/(ZL + Z0) and Vi(2B) = Vi*exp(-2*i*B), and some algebra, you arrive to the sin/cos formula you used in the original post. Between the effects of delay B and reflection coefficient G, the equivalent impedance at the source can be a very different value from both Z0 and ZL.