[Miguel045]

If I assume that Current (I) and voltage (V) are complex-valued quantities. How can I prove that Omh's law is indeed true? That is,

IZ=V

Where Z is the impedance!

I have no clue as of how to approach this problem. Any help will really be appreciated.
Thank you.

[harum]

Ohm's law for alternative current is fundamentally no different from DC Ohm's law in the sense that both laws state that the current is proportional to the applied voltage. If you really want to derive it, consider a circuit containing resistor (R), capacitor (C) and inductor (L), connected in sequence to a power supply V=Vo*exp(iwt), then:

Vo*exp(iwt) = L* d(dq/dt) /dt + R*dq/dt + q/C, here each term in the sum represents the drop in the voltage at each element, q is charge flowing through the device, and d(... )/dt - differentiation wrt time, w - power supply circular frequency. Solve this equation for q (for which you have to look up linear, non-uniform differential equations), differentiate it for time to get current from charge (I=dq/dt).

You will end up with an expression like this:

I = V / (R+ I*(wL - 1/wC)), where w, R, L and C are real. If you call the denominator Z, then V=IZ.

I don't know if this helps, but this is pretty much how you can derive it from scratch.

[julioisapantera]

Next is the easiest proof that I know.
I will do it for a capacitor of capacitance C, for which the current I(t) and the voltage v(t) are related by I(t)=Cdv(t)/dt

Take Fourier transform of that equation: I(jw)=CjwV(jw)
Define as impedance Z the ratio V/ I and find out that V(jw)/ I(jw)=Z=1/jwC
Done!

Can do similar proofs for R in series with C
R in series with L, etc

The real challenge is the physical interpretation of above math.