= Solution
Choose the <phasor> convention $V(t)=\operatorname{Re}(\widetilde V e^{i\omega t})$ and similarly for the <electric current>. In a linear sinusoidal steady state, differentiation becomes multiplication by $i\omega$, and <complex impedance> is $Z=\widetilde V/\widetilde I$. <Kirchhoff's current law> and <Kirchhoff's voltage law> then become algebraic equations that retain both amplitude and <phase> information.
For a <resistor>, $V=RI$, so $Z_R=R$. For a <capacitor>, $I=C\dot V$, while an <inductor> obeys $V=L\dot I$. Therefore
$$
\boxed{Z_R=R,\qquad Z_C=\frac1{i\omega C}=-\frac{i}{\omega C},\qquad Z_L=i\omega L}.
$$
The resistor's <electric potential difference> and current are in <phase>. The inductor's <electric potential difference> leads its current by $\pi/2$; the capacitor's current leads its <electric potential difference> by $\pi/2$. Replacing $e^{i\omega t}$ by $e^{-i\omega t}$ conjugates every impedance but leaves these physical lead-and-lag statements unchanged. The <phasor> method describes the steady state; switch-on transients must be added separately when initial conditions are relevant.
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