I’m a little confused on how complex numbers are used in circuit analysis . On one hand, complex numbers can only be used in linear circuits which do not involve any squares, square roots, and so forth of the voltage or current, or multiplication of one voltage or current by another. This way the real and imaginary parts don’t get mixed up. On the other hand, we do divide complex numbers by each other (e.g. when calculating complex impedance we divide voltage (in complex form) by current (also in complex form)) and the numbers still don’t get messed up. Why is it so? Any help will be much appreciated.

  • \$\begingroup\$ en.wikipedia.org/wiki/Phasor \$\endgroup\$ Apr 25, 2014 at 23:10
  • 1
    \$\begingroup\$ Complex numbers do get squared (and higher). Look at filter transfer functions for instance. \$\endgroup\$
    – Kaz
    Apr 26, 2014 at 0:36
  • \$\begingroup\$ I have never heard of the rules you talk about, why can't complex numbers be squared, square rooted e.t.c ? \$\endgroup\$
    – KillaKem
    Apr 26, 2014 at 4:31
  • \$\begingroup\$ the info that complex numbers can only be used in linear circuits was taken from this article (see page 4) scipp.ucsc.edu/~johnson/phys160/ComplexNumbers.pdf \$\endgroup\$
    – RuslanM
    Apr 26, 2014 at 5:46
  • \$\begingroup\$ @Kaz no, if you represent a 1Hz 1V signal as the complex phasor 1+0j, then feed in into a circuit which squares it, resulting in a signal with twice the frequency and a DC offset. This has no relation to the square of the phasor, (1+0j)² which is also (1+0j). Filters don't raise the phasor to a power, but can combine multiple stages ( so the complex gain is the product of multiple gains, which is OK - the question of why the gain is OK but not the power raising is a good one ). \$\endgroup\$ Apr 26, 2014 at 9:51

3 Answers 3


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.

  • \$\begingroup\$ Thanks for all the comments, it makes much more sense now. The impedance and power are not phasors but just complex numbers so it is still valid to divide voltage by current or multiply I by R^2. But then we also use those complex numbers (e.g. impedance) to calculate phasors (V and I). For example, if we have a circuit with 2 elements in series then V1=Z1/(Z1+Z2)*Vs and so on. Could somenone point me out to a formal proof that the real part of the phasor (amplitude and phase change) don’t get messed up because of these operations. Thank you again! \$\endgroup\$
    – RuslanM
    Apr 27, 2014 at 6:50
  • \$\begingroup\$ @user40908 the f(x) part of my answer does that - for a given frequency, capacitors and inductors work with the differential and integral of the cos omega t signal, which are w sin wt and 1/w sin wt, which is F cos (omega t + phi) for suitable values of F and phi. So any passive circuit can be reduced to a phasor scaling operation and phase change, the rest of the maths as below. \$\endgroup\$ Apr 27, 2014 at 15:38

You can represent a pure AC signal $$\begin{eqnarray*} v(t) = V \cos (\omega t + \theta) \end{eqnarray*}$$ using Euler's formula $$\begin{eqnarray*} e^{jx} = \cos x + j\sin x \end{eqnarray*}$$ by the real part
$$\begin{eqnarray*} v(t) = real ( V ( \cos (\omega t + \theta) + j\sin (\omega t + \theta) )\\ = real ( V e ^ {j(\omega t + \theta) }) \end{eqnarray*}$$ The phasor representation of \$v(t)\$ assumes a value of \$\omega\$ allowing us to write $$\begin{eqnarray*} v_{phasor} = V e ^ {j \theta} \end{eqnarray*}$$

The function of a linear circuit \$f\$ can be represented by a given gain and phase change

$$\begin{eqnarray} f(v(t)) = F V \cos (\omega t + \theta + \phi)\\ = real ( F V ( \cos (\omega t + \theta + \phi) + j\sin (\omega t + \theta + \phi) )\\ = real ( F V e ^ {j(\omega t + \theta + \phi) })\\ = real ( V e ^ {j(\omega t + \theta)} e ^ {j \phi })\\ = real ( A V e ^ {j(\omega t + \theta)} ) \end{eqnarray}$$ where \$A=F e ^ {j \phi }\$ is a complex gain for the circuit, the effect of the circuit on the time varying voltage is the same as calculating the product of the gain on the phasor representation.

So it can been seen that for a linear circuit, because the resulting time-varying waveform is still only a of \$\prod G . V \cos(\omega t + \sum \phi + \theta) \$ terms it has a valid phasor representation.

If instead of a linear function we have a non-linear function \$q\$ such as \$q(u)=u^2\$ : $$\begin{eqnarray} q(v(t)) = V^2 \cos^2 (\omega t + \theta )\\ = V^2\frac{1+ \cos 2(\omega t + \theta )}{2}\\ = real ( \frac{V^2}{2} (1+ e ^ {j(2\omega t + 2 \theta) })))\\ = \frac{V^2}{2} + real ( \frac{V^2}{2} e ^ {j(2\omega t + 2\theta) })\\ \end{eqnarray}$$

Now we no longer have a waveform which is a function of cos and sine \$\omega t\$ but has both DC and AC at \$2\omega t\$ . Because the frequency of this waveform is different to the one assumed by the original phasor, there is no phasor equation which can join them together - the phasor representation simplifies the equations by eliminating the common \$\omega t\$ variation, and this only works if you have the same frequency for each value in the equation.

If you can partition your system into linear sub-blocks then you can use the phasors within each block, taking care to convert the frequency at each non-linear function, but only if you have pure sinusoids at each stage. Other non-linear functions will result in other frequencies being produced, or non-sinusoidal waveforms. It is possible that these are represented by a complex FFT, and if that is fed into a linear system then each part of the FFT can be analysed using phasor analysis.

But the simple phasor relationships won't work once you have non-linearity, as all non-linearity distorts the sinusoid, and any distortion introduces additional frequencies, breaking the core assumption of the phasor representation.


The reason why complex numbers show up when doing only linear calculations is because the computations involve matrices. All systems of equations correspond to matrix equations and and all matrix equations correspond to systems of equations.

Now if you've taken a linear algebra course you should have encountered the notion of a change of basis. Essentially, it allows you to realize there was some choice in the way you wrote your system of equations and you could have written another system of equations which both describe the same physical situation.

What would be great is if through some clever reorganizing of our system of equations we ended up with a diagonal matrix of the form $$\begin{bmatrix} \lambda_1 & 0 & ... & 0\\ 0 & \lambda_2 & ... & 0\\ ... & ... & ... & ...\\ 0 & 0 &... & \lambda_n\\ \end{bmatrix}$$ because we understand solutions to any system of equations where this matrix is the coefficient matrix. That is systems of equations equivalent to $$\begin{eqnarray*} \lambda_1 x_1 &=& a_1\\ \lambda_2 x_2 &=& a_2\\ ... &=& ...\\ \lambda_n x_n &=& a_n\\ \end{eqnarray*}$$

which obviously has the solution \$x_k = \frac{a_k}{\lambda_k}\$ as long as \$\lambda_k \neq 0\$.

Now this isn't possible strictly speaking without the notion of generalized eigenspaces because of bad matrices such as $$\begin{bmatrix} 1 & 1 \\ 0 & 1\\ \end{bmatrix}.$$

However it does work sometimes. Consider the matrix $$\begin{bmatrix} -1 & 2\\ -1 & 1\\ \end{bmatrix}$$ becomes $$\begin{bmatrix} j & 0\\ 0 & -j\\ \end{bmatrix}$$ after a clever change of basis.

This is not just a silly use of \$j\$ for the sake of using \$j\$. If you want to write this matrix in diagonal form then you must involve the imaginary number \$j\$.

More concretely, this says that the system of equations with real coefficients $$-x + 2y = a$$ $$-x + y = b $$ is the same as the system of equations with complex coefficients $$jx = a'$$ $$-jy = b'.$$

The second system is clearly much easier to solve than the first but at the expense of involving \$j\$.

Even though I said this doesn't work exactly like this in general, even in the more general case you still will end up needing to consider complex numbers.

Edit: Other users have mentioned phase shift / phase analysis. The complex numbers in the phase shift come from this same fact. If you consider the two dimensional vector space spanned by \$\cos(t)\$ and \$\sin(t)\$ then the matrix representation of the 90 degree phase shift linear operator \$f(t) \mapsto f(t + \pi/2)\$ is given by the matrix $$\begin{bmatrix} 0 & -1\\ 1 & 0 \\ \end{bmatrix}$$ which is not diagonal but is equivalent to the diagonal matrix $$\begin{bmatrix} j & 0\\ 0 & -j\\ \end{bmatrix}.$$

This is the reason that complex numbers show up in AC circuit analysis.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.