The current through a capacitor is given by \$i_c=C\cdot \frac{dv_c}{dt} \$.
Let's say the voltage across the capacitor is a cosine wave. \$v_c=\cos(\omega t) \$.
Due to the complex exponential function we can write this as \$v_c=\Re(e^{j\omega t}) \$.
Let's calculate the current
$$i_c=C \cdot \frac{d}{dt}\Re(e^{j\omega t}) $$
$$i_c=C \cdot j\omega \Re(e^{j\omega t}) $$
$$i_c=j\omega C\cdot v_c $$
The impedance is defined as \$Z_c=\frac{v_c}{i_c} \$.
Which finally makes us arrive at
$$Z_c=\bigg(\frac{i_c}{v_c}\bigg)^{-1}=\frac{v_c}{i_c}=\frac{1}{j\omega C} $$
A similar thing can be done for an inductor.
My question is, do these formulas always hold? In the derivation I assumed that the voltage was a sinusoid (well, a phase shifted sinusoid) but this is not always the case.
What if the voltage across the capacitor is a sawtooth function or maybe a triangle wave? Then the derivation above wouldn't work at all.
Best Answer
The differential equations which use \$\frac{di}{dt}\$ and \$\frac{dV}{dt}\$ are more fundamental. They do not do not care about any abstractions such as "frequency", "sinusoids", or "canned" waveforms which, in a sense, need you to know what is going to happen in the future. As a result, you can always use the differential equations.
The impedance equations which use \$\omega\$ are derived from the differential equations using sinusoids as inputs. If you choose to work with the impedance equations instead of the differential equations then you need to break the input down into component sinusoids using Fourier analysis, perform the analysis for each sinusoid than add them up again at the end via super position. Don't forget to account for the phase shifts.