22Note that we begin this analysis with an exponential expression of the current waveform rather than the voltage waveform as we did at the beginning of the capacitor analysis. It is possible to begin with voltage as a function of time and use calculus to determine current through the inductor, but unfortunately that would necessitate integration rather than differentiation. Differentiation is a simpler process, which is why this approach was chosen. If ejωt = LdI dt then ejωt dt = LdI. Integrating both sides of the equation yields ejωt dt = L dI. Solving for I yields ejωt jωL plus a constant of integration representing a DC component of current that may or may not be zero depending on where the impressed voltage sinusoid begins in time. Solving for Z = V∕I finally gives the result we’re looking for: jωL. Ugly, no?