@Hemant, I feel that you might have been slightly left behind and that perhaps some of the discussion here and some of my earlier comments as well were not helpful for building your understanding. I'd like to try and summarise, just in case you're still a bit confused.
First consider an oscillating voltage ##v = v_0 \cos{(\omega t + \phi_0)}## applied to either a resistor, capacitor or inductor. Through the component a current will flow, oscillating each time at the same angular frequency but with perhaps an additional phase ##\varphi##, that is ##i = i_0 \cos{(\omega t + \phi_0 + \varphi)}##. Let's consider the AC response for resistors, capacitors and inductors in turn:$$\begin{align*}
\mathrm{R}: &\, i = \frac{v}{R} = \frac{v_0}{R} \cos{(\omega t + \phi_0)} \\ \\
\mathrm{C}: &\, i = \frac{dq}{dt} = C \frac{dv}{dt} = -\omega C v_0 \sin{(\omega t + \phi_0)} = \omega C v_0 \cos \left(\omega t + \phi_0 + \frac{\pi}{2} \right) \\ \\
\mathrm{I}: &\, i = i(0) + \int_0^{t} \frac{di}{d\xi} d\xi = i(0) + \int_0^t \frac{v_0}{L} \cos{(\omega \xi + \phi_0)} d\xi = \frac{v_0}{\omega L} \sin{(\omega t + \phi_0)} = \frac{v_0}{\omega L} \cos \left( \omega t + \phi_0 - \frac{\pi}{2} \right)
\end{align*}$$You may notice that the resistor response is in phase with the voltage, whilst the capacitor and inductor responses are ##\pi / 2## ahead and behind the voltage respectively.
It is convenient to introduce the
complex voltage and
complex current, defined by ##V = V_0 e^{i \omega t}## and ##I = I_0 e^{i \omega t}## respectively, where ##V_0 = v_0 e^{i \phi_0}## and ##I_0 = i_0 e^{i (\phi_0 + \varphi)}## are
complex amplitudes encoding the magnitude and initial phase. The ##\mathrm{Re}## function maps from the complex voltage/current and standard voltage/current, that is, ##v = \mathrm{Re}(V)## and ##i = \mathrm{Re}(I)##. One helpful property to remember is that the ##\mathrm{Re}## function commutes with time-differentiation; given some ##z: \mathbb{R} \rightarrow \mathbb{C}## with ##t \mapsto z(t)##, you can write ##z(t) = a(t) + ib(t)## and thus$$\mathrm{Re} \left( \frac{\mathrm{d}z}{\mathrm{d}t} \right) = \mathrm{Re} \left( \frac{\mathrm{d}a}{\mathrm{d}t} + i \frac{\mathrm{d}b}{\mathrm{d}t} \right) = \frac{\mathrm{d}a}{\mathrm{d}t} = \frac{\mathrm{d}\left( \mathrm{Re}(z) \right)}{\mathrm{d}t} $$What this means is that you can do all the differentiations in complex numbers, and convert everything back only at the very end.
Now, consider a "black box" consisting of some network of resistors, inductors and capacitors. It may be assigned a (complex)
impedance ##Z##, which you may think of loosely as a complex analogue of resistance, and is defined as the ratio ##Z := V / I##. For example, for the resistor you simply have ##I = V/R## and thus$$Z_R = \frac{V}{I} = R$$then for the capacitor, you have ##I = C \frac{dV}{dt} = i\omega C V_0 e^{i \omega t} = i\omega C V## and thus$$Z_C = \frac{V}{I} = \frac{1}{i\omega C }$$and for the inductor, you have ##V = L \frac{dI}{dt} = i\omega L I_0 e^{i \omega t} = i\omega L I## and thus$$Z_I = \frac{V}{I} = i \omega L$$How do you combine impedances? Let's just do one example, e.g. say ##N## impedances ##\{Z_1, \dots, Z_N \}## in
parallel. The complex voltage ##V## across each of these is the same, and the sum of the complex currents through each must equal the total current into the parallel network, hence$$\sum_{k=1}^{N} I_k = V\sum_{k=1}^{N} \frac{1}{Z_k} = \frac{V}{Z_{\mathrm{eff}}}$$which upon division by ##V## yields the familiar formula.
You might wonder, do all of Kirchoff's laws, and the rules you know for analysing circuits hold also on the complex amplitudes ##V_0## and ##I_0##? Yes! For example, consider the current law applied to a junction with ##N## inputs, $$(\forall t \in D) \quad \mathrm{Re}\left( \sum_{k=1}^N (I_0)_k e^{i \omega t} \right) \implies \sum_{k=1}^{N} (I_0)_k = 0$$because the only way for the sum to be zero for all values of ##t## is if the sum of the complex current amplitudes themselves vanish. The same analysis holds for the voltage law.
Why is this formalism useful? Well, let's consider your example of a circuit containing a resistor, capacitor and inductor in series, connected across an oscillating voltage source. You already know from the "impedances in series" principle that the resulting impedance is$$Z = Z_R + Z_L + Z_C = R + i \left(\omega L - \frac{1}{\omega C} \right)$$From this it's not hard to write down the current,$$I = \frac{V}{Z} = \frac{V}{R + i \left(\omega L - \frac{1}{\omega C} \right)} = \frac{V e^{- i \phi_Z}}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}}$$where the phase shift ##\phi_Z## is given by$$\phi_Z = \arctan \left( \frac{\omega L - \frac{1}{\omega C}}{R} \right)$$With this we can now work out some interesting quantities. For instance, what is the rate of energy dissipation in this oscillator, due to the resistor? You have$$\langle P \rangle = R|I_0|^2 \langle \cos^2{(\omega t - \phi_Z)} \rangle = \frac{1}{2}R |I_0|^2 = \frac{R|V_0|^2}{2(R^2 + (\omega L - \frac{1}{\omega C})^2)}$$At resonance, ##\omega_0 L - \frac{1}{\omega_0 C} = 0##.
Half-power thus occurs when ##\omega_h L - \frac{1}{\omega_h C} = \pm R##, or in other words when$$\omega_h = \mp \Gamma + \sqrt{\Gamma^2 + \omega_0^2}$$with ##\Gamma := R/2L##. Then, you may identify the full-width half-power as ##\delta \omega = \frac{R}{L}##, which results in a
quality factor ##Q = \frac{\omega_0}{\delta \omega} = \frac{1}{R} \sqrt{\frac{L}{C}}##.
In other words, complex numbers allow you to fairly easily analyse key properties of oscillating circuits, and oscillating systems in general.
sorry.
