Electromotive force, abbreviated as E.M.F and denoted by $\varepsilon$, is not a force. It is defined as the energy utilized in assembling a charge on the electrode of a battery when the circuit is open.Simply, it is the work done per unit charge which is the potential difference between the electrodes of the battery measured in volts. Mathematically, $\textbf{V} = \frac{\textbf{W}}{\textbf{q}}$.
Initially, energy is available in the form of chemical energy. This energy is utilized to take a charge say $+q$ to the anode by overcoming the electrostatic force of attraction due to the the negative charges on the cathode and the electrostatic force of repulsion due to the positive charges on the anode. The chemical energy then gets transformed into electrostatic potential energy present in the electric field between the electrodes of the battery.
Note that the sum of two phasors is another phasor:
$Ae^{j(wt+\phi_1)} + Be^{j(wt+\phi_2)} = Ce^{j(wt+\phi_3)}$
Where A,B,C are real.
The only way for the real part of the right side to be equal to zero at all times is if $C=0$. In which case the whole thing is $0$, both the real and complex parts are 0.
So a sum of phasors (of the full sinusoidal form with time) with a zero real part at all times, must have a zero complex part.
EDIT: Note that here by phasor, I mean the full sinusoidal form with the time included: $Ae^{j(wt+phi_1)}$, not necessarily the shortened $Ae^{j\phi_1}$.
To be clear, we can have:
$Ae^{j\phi_1} + Be^{j\phi_2} = Ce^{j 90^\circ}$
where C is nonzero and in this situation we have a $0$ real part, and a nonzero complex part on the right hand side. This doesn't apply to our situation, because we're talking about time-signals that are 0 at all times. If I take $Ce^{j 90^\circ}$ and convert it to the time domain, I get $Ccos(wt+90^\circ)$ where C is nonzero. This isn't 0 at all times as we require. Only way to get $0$ at all times is if $C=0$.
The thing with the kirchoff laws is that they give $0$ for all time.
To put it simpler:
If some time signal $v(t)$ is $0$ for all t, then its phasor form must be $0$ (complex and real parts are both 0).
Another way to look at it:
Phasor transforms are linear and one-to-one. So a $v_2(t) = k v_1(t)$, where k is some real number, then in the phasor domain $V_2 = kV_1$. so if $k=0$ we immediately get a 0 phasor in the phasor domain for $v_2$.
Best Answer
If you have an LCR series circuit connected to an alternating voltage supply then at an instant of time the current through each component in the circuit is the same and the variation of current with time is represented by the top graph in the diagram below.
The addition of voltages in the circuit is complicated that they do not reach a maximum value at the same time.
If you study the graphs above you will see that the voltage across the capacitor lags behind the voltage across the resistor by a quarter of a period which is equivalent to a phase angle of $-90^\circ$ and the voltage across the inductor leads the voltage across the resistor by a quarter of a period which is equivalent to a phase angle of $+90^\circ$.
So to find the voltage across all three of the components $v_{\rm series}$ at any instant of time one has to evaluate $v_{\rm R} \sin (\omega t) + v_{\rm C} \sin (\omega t - 90^\circ) + v_{\rm L} \sin (\omega t + 90^\circ)$.
A convenient way to do this addition is to use a phasor diagram as shown on the right and you will see in this case $v_{\rm supply}$ leads the voltage across the resistor (and hence the current which is always in phase with the voltage across the resistor by a value which is between $0^\circ$ and $+90^\circ$.