Power in complex notation

Every AC power triangle you have ever drawn — active power along one leg, reactive power along the other, apparent power as the hypotenuse — is secretly a single complex number in disguise. Complex power packages $P$ , $Q$ and $S$ into one algebraic object, $S = VI^{*}$ , built directly from the voltage and current phasors you already compute for any AC circuit. No geometry, no separate triangle sketch: multiply two phasors (with one conjugated), and the real part, imaginary part and magnitude hand you $P$ , $Q$ and $|S|$ in one step. This article shows why the conjugate is essential, derives the rectangular split, and gives a fast alternative route through impedance, $S = I^2Z$ , for when the load impedance is already known.

The core insight

Complex power $S = VI^{*}$ — voltage phasor times the conjugate of the current phasor — packs active power $P$ (real part), reactive power $Q$ (imaginary part) and apparent power $|S|$ (magnitude) into one complex number computed straight from phasor algebra.

The core idea

Complex power is defined from the RMS voltage and current phasors of a single-phase circuit:

S = VI^{*}

where: $V$ is the RMS voltage phasor, $I$ is the RMS current phasor, and $I^{*}$ is its complex conjugate — same magnitude, angle negated. Conjugating the current (never the voltage) is what makes the sign of the imaginary part come out right for both lagging and leading loads. Writing $V = |V|\angle\alpha$ and $I = |I|\angle\beta$ , the conjugate is $I^{*} = |I|\angle(-\beta)$ , so multiplying phasors (magnitudes multiply, angles add) gives

S = |V||I|\angle(\alpha-\beta) = |V||I|\angle\theta

where: $\theta = \alpha - \beta$ is exactly the voltage-current phase angle already used to define power factor, $\cos\theta$ . Expanding this polar form into rectangular (real + imaginary) form separates $S$ into the two familiar quantities:

S = |V||I|\cos\theta + j\,|V||I|\sin\theta = P + jQ

where: $P = VI\cos\theta$ is the active power in watts (the real part of $S$ ), $Q = VI\sin\theta$ is the reactive power in VAR (the imaginary part), and $|S| = VI$ is the apparent power in VA (the magnitude of $S$ ). These are exactly the power-triangle relations from the impedance-based treatment, now falling out of phasor multiplication rather than a geometric drawing. The sign of $Q$ carries physical meaning: a positive $Q$ ( $\theta > 0$ , current lagging voltage) marks an inductive load absorbing reactive power, while a negative $Q$ (current leading) marks a capacitive load supplying it — matching the impedance convention $Z = R + jX$ with $X = X_L - X_C$ . When the load impedance $Z$ is already known, there is a faster route that avoids re-finding the voltage phasor: since $V = IZ$ for a load carrying current $I$ , substituting gives

S = VI^{*} = (IZ)I^{*} = |I|^2 Z

where: $|I|^2$ is the square of the current magnitude (not conjugated), and $Z$ is the complex load impedance. This is the quickest path to $S$ whenever $R$ and $X$ are already on hand.

Figure

A power triangle with active power along the horizontal leg, reactive power along the vertical leg, apparent power as the hypotenuse at angle theta, labelled with the complex power S equals P plus jQ

Derivation ASplitting complex power into active and reactive power from phasor magnitudes and angles

The goal is to show, starting from nothing more than the definition $S = VI^{*}$ and the ordinary rules for multiplying phasors, exactly why the real part of complex power is the active power $P = VI\cos\theta$ and the imaginary part is the reactive power $Q = VI\sin\theta$ , so that the familiar power-triangle numbers can be read straight off a single complex multiplication instead of a separate geometric construction.

Step 1. Write the voltage and current phasors in polar form: $V = |V|\angle\alpha$ and $I = |I|\angle\beta$ , where $|V|$ and $|I|$ are RMS magnitudes and $\alpha$ , $\beta$ are their phase angles relative to a common reference.

Step 2. Conjugate the current. Conjugation keeps the magnitude and negates the angle, so $I^{*} = |I|\angle(-\beta)$ . This is the one non-obvious move in the whole derivation, and it is what fixes the sign convention for $Q$ .

Step 3. Multiply the phasors. For phasors, multiplication multiplies magnitudes and adds angles, so $S = VI^{*} = |V||I|\angle(\alpha - \beta)$ . Defining $\theta = \alpha - \beta$ as the voltage-current phase angle, this is $S = |V||I|\angle\theta$ , a single complex number of magnitude $VI$ at angle $\theta$ .

Step 4. Convert to rectangular form using Euler's relation, $\angle\theta \equiv \cos\theta + j\sin\theta$ , giving $S = VI\cos\theta + j\,VI\sin\theta$ .

Step 5. Read off the parts. The real part is $VI\cos\theta$ ; since $\cos\theta$ is the power factor, this term is exactly the active power $P$ already defined from the power triangle. The imaginary part is $VI\sin\theta$ , the reactive power $Q$ . The magnitude $|S| = \sqrt{P^2+Q^2} = VI$ is the apparent power, and the angle recovered from $S$ , $\theta = \tan^{-1}(Q/P)$ , is the same phase angle used everywhere else.

Worked example

A load is supplied by $V = 200\angle0^\circ$ V and draws current $I = 10\angle{-30^\circ}$ A. Find the complex power and identify $P$ , $Q$ , $|S|$ and the power factor, then cross-check using $S = I^2Z$ .

Step 1. Conjugate the current: $I^{*} = 10\angle30^\circ$ A (magnitude unchanged, angle negated).

Step 2. Multiply phasors: $S = VI^{*} = (200\angle0^\circ)(10\angle30^\circ) = 2000\angle30^\circ\ \text{VA}$ .

Step 3. Convert to rectangular form: $S = 2000\cos30^\circ + j\,2000\sin30^\circ = 2000(0.866) + j\,2000(0.5) = 1732.1 + j1000\ \text{VA}$ .

Step 4. Read off the results: $P = 1732.1\ \text{W}$ , $Q = 1000\ \text{VAR}$ (positive, so the load is inductive and lagging), $|S| = 2000\ \text{VA}$ , and $\cos\theta = \cos30^\circ = 0.866$ lagging, since the current lags the voltage by $30^\circ$ .

Step 5 — cross-check with impedance. The load impedance is $Z = V/I = (200\angle0^\circ)/(10\angle{-30^\circ}) = 20\angle30^\circ\ \Omega$ . Using $S = I^2Z$ with the current magnitude squared, $S = 10^2 \times 20\angle30^\circ = 2000\angle30^\circ\ \text{VA}$ , exactly matching Step 2. Final answer: $S = 1732.1 + j1000\ \text{VA}$ , so $P = 1732.1\ \text{W}$ , $Q = 1000\ \text{VAR}$ , $|S| = 2000\ \text{VA}$ , power factor $0.866$ lagging.

Common traps

Using $S = VI$ without conjugating the current — this flips the sign of $Q$ whenever the phase angle is negative; always conjugate $I$ , never $V$ .
Getting a negative computed $P$ for a passive R-L-C load — since active power must be non-negative there, a negative result signals a conjugation or sign error, not a real physical answer.
Confusing units — $P$ is in watts, $Q$ in VAR, $S$ (or $|S|$ ) in VA; all three are dimensionally volt-amperes but are named differently to show their physical role.
Applying $S = I^2Z$ with the wrong current — it must be the current through that specific impedance, not the total supply current, once a circuit is not a simple series loop.

Ask the tutor: if the conjugation step in $S=VI^{*}$ still feels unmotivated, ask the AI tutor to derive it from first principles starting with instantaneous power $p(t) = v(t)i(t)$ .

Why do we conjugate the current and not the voltage in S = VI*?

The conjugate is what makes the sign of $Q$ track the physical behaviour of the load correctly. If you instead conjugated the voltage (or conjugated neither), a lagging (inductive) load and a leading (capacitive) load would not come out with opposite-signed reactive power in a way that matches the impedance convention $X = X_L - X_C$ . Conjugating $I$ is simply the definition that keeps complex power consistent with the power-triangle sign convention you already use.

Is S = I^2 Z the same thing as S = VI*, or a different formula?

It is the same formula, just substituted differently. Since $V = IZ$ for a load with impedance $Z$ , plugging into $S = VI^{*}$ gives $S = (IZ)I^{*} = |I|^2 Z$ , because $I \cdot I^{*}$ is just the magnitude squared. Use whichever form matches the data you are given — voltage and current phasors call for $S = VI^{*}$ ; a known impedance and current magnitude call for $S = I^2Z$ , and both must agree, as the worked example shows.

Why does the sign of Q tell us inductive versus capacitive, and not P?

$P$ , the real part of $S$ , is the active power actually converted to useful work or heat — for any passive load it is always positive, so its sign carries no information about the load type. $Q$ , the imaginary part, tracks energy that sloshes back and forth without being consumed, and its sign directly mirrors the reactance: positive $Q$ pairs with positive (inductive) reactance and lagging current, negative $Q$ pairs with negative (capacitive) reactance and leading current. That is why only $Q$ 's sign is used to classify the load.

Power in complex notation

The core idea

Worked example

Common traps

Before this

Where this leads