Parallel RLC circuit

There have been a lot of comments but I think it's still valuable to sketch the solution:

The impedances of \$L\$ and \$C\$ are given at an angular frequency \$\omega = 2000\$ rad/s. This means that

$$\omega L = 6\Omega\text{ and } \frac{1}{\omega C}=8\Omega$$ which gives the following values for \$L\$ and \$C\$: $$L=3mH\quad C=62.5\mu F$$

If you don't know the formula for the resonance frequency by heart, it's very easy to derive it (if you know that the impedance must be purely real-valued at resonance):

$$Z = R + j\left (\omega L -\frac{1}{\omega C} \right)$$ The imaginary part of the impedance \$Z\$ disappears for \$\omega L = \frac{1}{\omega C}\$, i.e.

$$\omega_0 = \frac{1}{\sqrt{LC}} = 2309.4\text{ rad/s}$$

Since at resonance the impedance of the capacitor and the inductor are equal, the voltages across them must be equal. So the current through the inductor at resonance must be

$$\frac{4mV}{\omega_0 L} = 0.577\text{ mA}$$

Assuming this equation is correct: $$Z = \frac{R}{1+i \omega cR} + i \omega L$$

We multiply the first term's nominator and the denominator by the complex conjugate of the denominator: $$.. = \frac{R(1-i\omega c R)}{(1+i \omega cR)(1-i\omega c R)} + i \omega L$$ and thus getting rid of imaginary stuff in the denominator: $$.. = \frac{R(1-i\omega c R)}{(1+ \omega^2 c^2R^2)} + i \omega L$$ And this thing is easily separated to real and imaginary parts: $$.. = \frac{R}{1+ \omega^2 c^2R^2} + i \left(\omega L - \frac{\omega c R^2}{1+ \omega^2 c^2R^2} \right)$$