= Solution
Use $e^{i\omega t}$ <phasors>. The relation $V=L\dot I$ gives $Z_L=i\omega L$. The three branches form a <parallel RLC circuit> and share the source <voltage>. The source-current <phasor> and its real time dependence are
$$
\widetilde I_{\rm total}=V_0\left[\frac1R+i\left(\omega C-\frac1{\omega L}\right)\right],
$$
$$
I_{\rm total}(t)=V_0\left[\frac{\cos\omega t}{R}-\left(\omega C-\frac1{\omega L}\right)\sin\omega t\right].
$$
Consequently its <amplitude> is $V_0\sqrt{R^{-2}+(\omega C-1/(\omega L))^2}$ and its <phase> relative to the <voltage> is $\tan^{-1}[R(\omega C-1/(\omega L))]$. At low <frequency> it is inductive and lags toward $-\pi/2$; at high <frequency> it is capacitive and leads toward $+\pi/2$. The ideal <amplitude> diverges at either extreme and has a finite minimum between them; real component losses and source <electrical impedance> limit those asymptotes.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1-parallel-resonance.png]
{title=Original normalized source-current amplitude, phase and resistor-to-source-current ratio for an illustrative ideal parallel RLC circuit}
Zero <phase> requires cancellation of <electrical susceptances>:
$$
\boxed{\omega_{\rm res}=\frac1{\sqrt{LC}},\qquad|I_{\rm total}|_{\rm res}=\frac{V_0}{R}\quad\text{(minimum)}.}
$$
The <inductor> and <capacitor> currents are nonzero, equal in magnitude and opposite in <phase>, while the <resistor> supplies the in-phase source current. <energy> oscillates between the reactive elements; dissipation occurs in the shunt <resistor>. A source-current maximum would describe a different, series-resonant circuit.
The specified ratio is
$$
\frac{|I_R|}{|I_{\rm total}|}=\frac1{\sqrt{1+R^2(\omega C-1/(\omega L))^2}}.
$$
It exceeds $1/\sqrt2$ when $|\omega C-1/(\omega L)|<1/R$. Define $b=1/(RC)$ and $\omega_0=1/\sqrt{LC}$. The two positive endpoints are $\omega_\pm=[\sqrt{b^2+4\omega_0^2}\pm b]/2$, so
$$
\boxed{\delta\omega=\omega_+-\omega_-=\frac1{RC},\qquad Q=\frac{\omega_0}{\delta\omega}=R\sqrt{\frac CL}.}
$$
The endpoints have geometric mean $\omega_0$. Increasing the parallel resistance reduces dissipative loading and narrows this band, in contrast with increasing the resistance of a <series RLC circuit>. Equivalently, $Q=\omega_0U_{\max}/P_{\rm loss}$, with $U_{\max}=CV_0^2/2$ and $P_{\rm loss}=V_0^2/(2R)$. The <parallel RLC bandwidth> here belongs to the explicitly defined current ratio, not a half-maximum rule applied to a source-current peak that does not exist.
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