On the other hand, we do divide complex numbers by each other
That's true. The impedance of a circuit element is the phasor voltage divided the phasor current
$$Z = \frac{\vec V}{\vec I} $$
But note that the impedance is not a phasor - it does not represent the amplitude and phase of a sinusoid like the voltage and current phasors.
Similarly, the complex power is the product of the (rms) phasor voltage and (rms) conjugate phasor current
$$S = \tilde V \cdot \tilde I^*$$
And again, the complex power is not a phasor, it is just a complex number.
That fact is that products and ratios of phasors are not phasors. Thus, we can't apply phasor analysis to non-linear circuits.
For example, let a circuit element voltage be proportional to the current squared:
$$v = ki^2 $$
If the current is a sinusoid of frequency \$\omega\$, the voltage is a constant plus a sinusoid of frequency \$2\omega\$.
$$v = k(I\cos\omega t)^2 = \frac{kI^2}{2}(1 + \cos2\omega t)$$
But, for phasor analysis, we depend on the fact that all the circuit voltages and currents are of the same form, i.e., are sinusoids of the same frequency, only differing in amplitude and phase.
Moreover, as pointed earlier, the square of a phasor is not a phasor thus we cannot square the current phasor and hope to get a voltage phasor.
Best Answer
It's pretty straightforward. Normally you would see a ground symbol instead of 0V. The other voltages on the schematic are node voltages, and (for your purposes) they can be treated like ideal voltage sources connected between the node and ground.
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