Solution

a) The voltage drop across the resistor is just Ohm's law:

$$V_R = I_0 R cos \omegat$$

There is a charge on the capacitor:

$$q = V_C C= \int_{0}^{t} I(t) dt$$

$$V_C C = \frac{I_0}{\omega} sin(\omega t)$$

$$V_C = \frac{I_0}{\omega C} cos(\omega t - \frac{\pi}{2})$$

Finally, we write down the equation for the voltage across the coil:

$$V_L = L \frac{dI}{dt} = -LI_0 \omega sin(\omegat)$$

$$V_L = \frac{dI}{dt} = LI_0 \omega cos(\omegat + \frac{\pi}{2})$$

The phase shift as seen in the coil $+\frac{\pi}{2}$ depends on the current, in the capacitor $-\frac{\pi}{2}$, the resistance has no shift, the amplitudes are equal:

$$V_R = I_0 R$$
$$V_C = \frac{I_0}{\omega C}$$
$$V_L = LI_0 \omega$$

b)
To find the amplitude, it is enough to add up our voltages vector based on the diagram obtained in a:

$$V = I_0 \sqrt{R^2 + (\omegaL - \frac{1}{\omega C})^2}$$

Since we have a right triangle formed from vectors, the phase shift is easy to express in terms of a tangent.:

$$\varphi = arctg \frac{\omegaL - \frac{1}{\omega C}{R}$$