KVL in the frequency domain - Why is the sum of phasors 0 and not just the real part of the sum?

Question

In many proofs of Kirchhoff's voltage law for the frequency domain, the time-domain KVL is originally used and then the time-dependent part (\$e^{jwt}\$) is removed, leaving only the real part of the sum of phasors to equal zero, for example in the textbook "Electric Circuits" by Nilsson and Riedel:

$$v_1 + v_2 + ... + v_n = 0$$ $$\therefore Vm_1cos(wt+A_1)+Vm_2cos(wt+A_2) + ... + Vm_ncos(wt+A_n)=0$$ $$\therefore Real [(\vec V_1 + \vec V_2 + ... + \vec V_n)e^{jwt}]=0$$

Then, this "real part" notation is simply removed, stating that the sum of phasors is simply equal to zero.

$$\vec V_1 + \vec V_2 + ... + \vec V_n$$

Why is this the case? Where can this be proved?

Answer 1

The expression

$$\text{Re}\, [(\vec V_1 + \vec V_2 + ... + \vec V_n)e^{j\omega t}]=0$$

must hold for all values of \$t\$; this means that it needs to hold when \$e^{j\omega t} = 1\$ and when \$e^{j\omega t} = j\$ (as well as all linear combinations of the above); this is only satisfied when the sum of the complex phasors is zero (both real and imaginary parts)

Answer 2

A couple minutes after posing this question, I realized the answer.

The equation $$Real[(\vec V_1 + \vec V_2 + ... + \vec V_3)e^{jwt}]=0$$

has to be true for any value of \$ t \$. If we take t to be where \$ wt = \frac \pi 2 \$, the phasors will all be rotated 90 degrees in the complex plane. This means that for a phasor \$ \vec V_i = a+bi\$, its rotated value will now be \$-b + ai \$.

The only way for the real part of the sum of the rotated vectors multiplied by \$ e^{jwt} \$ to always be 0, is for both the imaginary parts to sum to zero and the real parts to sum to zero, since 90 degrees later the imaginary parts become the negative real parts etc.

Hope this makes sense.