Complex power and why VI* rather than V*I

Complex power is $S = P + jQ$, the phasor quantity whose real part is the active power (W), whose imaginary part is the reactive power (var) and whose magnitude is the apparent power (VA). It is defined as $S = VI^*$.

Why $VI^*$

Let $V = |V|\angle\theta_v$ and $I = |I|\angle\theta_i$. The power-factor angle is $\phi = \theta_v - \theta_i$.

$$\begin{aligned} VI^* &= |V||I|\angle(\theta_v - \theta_i) = |V||I|\cos\phi + j|V||I|\sin\phi = P + jQ\\ V^*I &= |V||I|\angle(\theta_i - \theta_v) = |V||I|\cos\phi - j|V||I|\sin\phi = P - jQ \end{aligned}$$

1. Both give the same P. Only the sign of Q differs.
2. Sign agreement with the accepted convention. With $VI^*$, a lagging (inductive) load ($\theta_i < \theta_v$, $\phi > 0$) gives $Q > 0$. Inductive loads "absorb" vars and capacitors "supply" vars. This is the convention adopted by IEEE/IEC and used in load-flow studies, generator capability charts and var meters. With $V^*I$, an inductive load would show negative Q, which conflicts with the convention.
3. Consistency with impedance. With $S = VI^* = |I|^2Z = |I|^2(R + jX)$, inductive reactance (+X) gives +Q. The power triangle is then the impedance triangle scaled by $|I|^2$. With $V^*I$ the triangle would be the mirror image of the impedance triangle.
4. Independence from reference. Taking the conjugate removes the absolute phase angles and keeps only the difference $\theta_v - \theta_i$. So S does not depend on which phasor is chosen as reference. ($VI$ without a conjugate would give the angle $\theta_v + \theta_i$, which changes with the reference and has no meaning.)

 Im(Q)
  ^        S = VI*
  |      /
  |    /  Q = |V||I| sin(phi) > 0
  |  / phi    (lagging load)
  +----------> Re (P)

Example: $V = 100\angle0^\circ$ V and $I = 10\angle-30^\circ$ A (lagging). Then $VI^* = 866 + j500$ VA, which is correct: the load absorbs 500 var. But $V^*I = 866 - j500$ VA would wrongly suggest that the inductive load supplies vars.