= Solution
Use sinusoidal <phasors> with physical values $v(t)=\operatorname{Re}(\widehat V e^{j\omega t})$, $i(t)=\operatorname{Re}(\widehat I e^{j\omega t})$, and $j^2=-1$. The <complex impedance> is $Z=\widehat V/\widehat I=X+jY$: its <modulus> is the voltage-to-current <amplitude> ratio, and its <complex argument> is the voltage-minus-current <phase difference>. Here $X$ is the <electrical resistance> part and $Y$ the signed <electrical reactance>; the source's $X,Y$ notation differs from the common $R,X$ notation.
The component laws $v_R=Ri$, $v_L=L\,di/dt$ and $i_C=C\,dv_C/dt$ give the <complex impedances> of an ideal <resistor>, <inductor> and <capacitor>:
$$
Z_R=R,\qquad Z_L=j\omega L,\qquad Z_C=\frac1{j\omega C}=-\frac j{\omega C}.
$$
In a <series circuit> the same <electric current> flows through both components, so <Kirchhoff's voltage law> adds their voltage <phasors>. The resulting <series RC impedance> is
$$
Z=R-\frac j{\omega C},\qquad
\boxed{|Z|=\sqrt{R^2+\frac1{\omega^2C^2}},\qquad
\arg Z=-\arctan\frac1{\omega CR}.}
$$
For positive $R,C,\omega$, the negative <phase difference> means that <electric current> leads the applied <electric potential difference>.
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