In a RLC circuit having a AC source . The actual current flowing in any branch will be the real part of the complex current. The imaginary component has no important role. It is just there. Now that the current is of the form $I_0 e^{i (\omega t + \phi)}$, we can use the formulas $V= IR$, $V= L \frac {dI}{dt}$ and $V= \frac Q C$ to find voltages across resistors, inductors and capacitors respectively. But why do we use $I=I_0 e^{i (\omega t + \phi)}$ for calculating voltages across inductors and capacitors? Shouldn't we only use $I=Re(I_0 e^{i (\omega t + \phi)})$ because only the real part of the complex current is the actual current flowing in the branch? If we use the real as well as the imaginary part of find the voltage across an inductor, then during differentiation the iota operator will "come down" and will be multiplied with the complex number and so the earlier-real part will become imaginary and the earlier-imaginary part will become real. Why is it okay to use complex current to find the voltage . And why we use $S= VI^*$, where $I^*$ conjugate of complex current. Why is it not $S=VI$ or $S=Re(V)Re(I)$ ?

  • $\begingroup$ This question was asked previously by someone but i think it needed a bit more rigorous explanation $\endgroup$ Commented Mar 10, 2022 at 12:08
  • $\begingroup$ "The imaginary component has no important role. It is just there." What makes you think that? $\endgroup$
    – Bob D
    Commented Mar 10, 2022 at 13:02
  • $\begingroup$ Hi ProblemDestroyer. Welcome to Phys.SE. Link to previously asked question? $\endgroup$
    – Qmechanic
    Commented Mar 10, 2022 at 16:19
  • $\begingroup$ @Qmechanic I do not have time to look at the moment but perhaps the words, complex real apparent power, might be a good place to start in the search engine? $\endgroup$
    – Farcher
    Commented Mar 10, 2022 at 16:31

3 Answers 3


The reason that the complex exponential is used, is because in deriving those formulas from the relevant differential equations, a common method of solving is by "substitution". The complex exponentials derivative is proportional to itself, meaning we can use substitution easily to find a relevant dispersion relation(as the proportionality constants cancel out). We could technically use the regular trig functions however those derivative are much harder to obtain any dispersion relation. Finally, we can prove that the re{} of the solution to the complex equation, is a solution to the non complex differential equation.

Another reason we might want to use complex numbers to represent certain quantities, is for compact notation.

Take ohms law for frequency dependant electric fields.

$$\vec{J} = re(\sigma \vec{E})$$

Here a complex value of $\sigma$ is used, which allows us to easily write the corresponding magnitude of the current density, and ALSO neatly encapsulating the corresponding phase difference between the applied electric field and current density. We could have easily chosen not to use complex numbers, however in this form, we can write it as a complex proportionality with the electric field, instead of some complicated trig function.

Given I want to derive ohms law for frequency dependant $\vec{E}$ fields.

Let $\vec{E} = \vec{E_{0}}cos(\omega t)$

From the drude model of conductivity, the equations of motion of an electron inside an ohms material follow:

$$m\frac{d\vec{v}}{dt}= q\vec{E}-\frac{m}{T}\vec{v}$$

Aka f= ma , with some driving force qE, and some resistive term proportional to v

Given we have defined $\vec{E}$, the differential equation we would like to solve is,

$$m\frac{d\vec{v}}{dt}= q\vec{E_{0}}cos(\omega t)-\frac{m}{T}\vec{v}$$

Now, consider instead, the differential equation:

$$m\frac{d\vec{v}}{dt}= q\vec{E_{0}}e^{i\omega t}-\frac{m}{T}\vec{v}$$

Let's call $v_{1}$ the solution to this differential equation,

Taking the real part of this equation gives

$re(m\frac{d\vec{v_{1}}}{dt}= q\vec{E_{0}}e^{i\omega t}-\frac{m}{T}\vec{v_{1}})$

$$m\frac{d}{dt}re(\vec{v_{1}})= q\vec{E_{0}}cos(\omega t)-\frac{m}{T}re(\vec{v_{1}})$$

Notice this is in the same form as the equation that we actually want to solve!, and the function that has been shown to satisfy this equation is $re(\vec{v_{1}})$

solving the complex equation is MUCH easier to do, and we have shown that the relation between solution the complex equation and the solution to the equation we want to solve, is that by taking the real part of the solution to the complex equation, gives us the solution to our original equation!

Which is why

$$\vec{J} = re(\sigma \vec{E})$$

And not. $$\vec{J} = re(\sigma) re( \vec{E})$$

  • $\begingroup$ I assume you really want to know about ohms law and why it is the real(), look up the drude model of conductivity $\endgroup$ Commented Mar 10, 2022 at 12:30
  • $\begingroup$ I'll edit my answer in a minute with some insight on specifically what I've said about ohms law $\endgroup$ Commented Mar 10, 2022 at 12:32

The imaginary component has no important role. It is just there.

That is not correct.

For example, if it were not for the imaginary component devices requiring magnetic fields, such as induction motors, wouldn't work. In order to convert electrical energy into mechanical energy a magnetic field needs to be created. It is the imaginary component of inductor current that creates the magnetic field.

It would appear that the rest of your post follows from an incorrect premise.

Hope this helps.


This is interesting question and is down to a convention which makes life easier.

Consider the complex forms of a voltage $Ve^{i\theta_{\rm V}}$ and current $Ie^{i\theta_{\rm I}}$ where $V$ and $I$ are the RMS values as shown in the phasor diagram below.

enter image description here

As you have noted the complex power is defines as the complex voltage times the complex conjugate of the current $S = Ve^{i\theta_{\rm V}}\times Ie^{-i\theta_{\rm I}}=VI e^{i(\theta_{\rm V}-\theta_{\rm I})}$.

You will now perhaps note that taking the complex conjugate of the current means that the complex power is $S = VI e^{i\phi}$ where $\phi= \theta_{\rm V}-\theta_{\rm I}$ is the phase angle between the voltage and the current and that angle does not depend on any initially defined imaginary and real axes.
And this is the short answer to your question.

The complex power $S$ measured in volt-amperes (VA) can now be split into a real power $P=VI \cos \phi$ measured in watts (W) and an apparent power $Q=VI \sin \phi$ measured in volt-amperes (VA).
The diagram I have drawn on the left is called the power triangle.

You may think that apparent power is not worth bothering about because it relates to components like capacitors and inductors which over a period consume no electrical power.
For example a capacitor is charged up taking energy from a voltage source and then is discharged returning the energy back to the voltage source.
On simple electrical meters which monitor electricity usage such charging and discharging would not be measured.
However consider what is actually happening.
When a capacitor is charged and then discharged a current flows though the connecting leads which have resistance.
Any leads dissipating electrical power which is not measured by a home/factory electric meter means that the company generating electricity are not charging you for the electrical energy dissipated as heat in the transmission lines due to you using devices which are not purely resistive
Apparent power becomes significant when dealing with industrial sized devices.


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