I'm doing an assignment on circuit analysis with phasors and it's brought up a point of confusion for me on how Phasors convert to rectangular form.

My textbook defines phasors as $$v(t) = V_M\text{cos}(\omega t + \phi) = \text{Re}[V_Me^{j(\omega t + \phi)} ]$$

and says that they can be written in phasor notation as $$ V_M\angle \phi $$

That makes plenty of sense to me. However, in my homework, when I have to add phasors and convert them to rectangular, the solution has them being represented as

$$ V_M\angle \phi = V_M\text{cos}(\phi) + jV_M\text{sin}(\phi) $$

This difference seems contradictory to me. The book seems to imply that phasors are only expressing the real component of the function, but when it comes to problems, phasors are now both the real and imaginary components? I know I have some conceptual misunderstanding somewhere, but I just can't identify it.

  • \$\begingroup\$ Remember that a complex number is just a way to represent a point in two dimensions, a phasor is another way. \$\endgroup\$
    – Mattman944
    Commented Mar 29, 2020 at 23:09
  • \$\begingroup\$ Have a look at the real and imaginary components of \$V_M\angle \phi\$. \$a = \text{Re}(V_M\angle \phi) = V_M\cos(\phi)\$ and \$b = \text{Im}(V_M\angle \phi) = V_M\sin(\phi)\$. Now \$V_M\angle \phi = a+jb\$. You now have the polar form with components \$V_M\$ and \$\phi\$, and the rectangular form with components \$a\$ and \$b\$. You use the rectangular form to add and subtract, and the polar form to divide and multiply. You can also use the rectangular form to multiply \$(a+jb)\cdot(c+jd) = (ac-bd)+j(ad+bc)\$. \$\endgroup\$
    – skvery
    Commented Mar 30, 2020 at 7:14
  • \$\begingroup\$ You might want to have a look at this answer of mine too. \$\endgroup\$ Commented Mar 30, 2020 at 7:43
  • \$\begingroup\$ It might be useful to recall the old observation "All models are wrong; some are useful". Including or excluding the imaginary component is ultimately a matter of practicality, not an absolute truth. \$\endgroup\$
    – MSalters
    Commented Mar 30, 2020 at 8:47
  • \$\begingroup\$ Thanks for all the added context, folks! \$\endgroup\$
    – wagboi
    Commented Mar 30, 2020 at 19:11

2 Answers 2


A phasor is a vector representation of a sinusoidal voltage (or current) frozen in time. Let's say the time domain equation is V(t) = A * cos(omega * t + phi). Please note that even though the wave is a sinusoid, there is magnitude (A) and an angle involved. For example, when you freeze the signal at time t = 0, then the sinusoidal voltage actually has an angle associated with it. In our case, that angle is phi (since t = 0). The magnitude is A.

Since the phasor has a magnitude and an angle or direction, it is actually a vector. When we convert it from polar coordinates to rectangular coordinates, we let one of the axes be the imaginary axis. That is how imaginary numbers come into the picture. When we freeze a rotating vector in time, the imaginary part provides information about the angle.

The instantaneous voltage is the real part of the vector. That is just how polar to rectangular coordinate conversions work.

You could, if you want, say that a sinusoidal voltage is really a snapshot of a mysterious rotating object that we can't see. It traces out a sine wave onto the real plane that we are capable of observing, but somewhere we can't see, there is another sine wave being traced out on an imaginary plane we cannot see. The imaginary sinusoid is out of phase by 90 degrees to the real one.

Later you may get into the concepts of in-phase and quadrature mixing of signals. Hilbert transforms and the Analytic Signal. It is all very interesting. My career didn't end up exposing me to that stuff in a deep way but it is abstract and hard to understand unless you spend some time struggling with it.

  • \$\begingroup\$ I see. So if I get some answer for a current or voltage and it's phasor, I just convert it into its real (or usually cosine) component, right? \$\endgroup\$
    – wagboi
    Commented Mar 30, 2020 at 3:14
  • \$\begingroup\$ Whatever is desired. Maybe sometimes the phasor answer could be acceptable. Maybe other times the time domain version is required. They contain the same information so either one SHOULD be ok, but if you are in school and taking a test, read the question carefully to see which one is requested. \$\endgroup\$
    – user57037
    Commented Mar 30, 2020 at 4:27
  • \$\begingroup\$ Yeah, the cosine is the real part and sin is the imaginary part. \$\endgroup\$
    – user57037
    Commented Mar 30, 2020 at 5:12
  • \$\begingroup\$ Got it, thanks! \$\endgroup\$
    – wagboi
    Commented Mar 30, 2020 at 5:52

There are two things to consider here.

The first is the purpose of phasor notation. I personally tend to think of it as "encoding" the steady-state time-varying information of a circuit into a single value (the phasor) that facilitates "simple" arithmetic. The first two equations in your question relate to this consideration: they tell you how to convert a single-frequency time-varying sinusoidal value into the simpler notation of a phasor. In essence, you're setting your analysis up to just use arithmetic and algebra to solve for voltages/currents in the circuit, just like you would with DC analysis. The only difference here is that you're dealing with complex numbers.

Which brings me to the second consideration. As you're dealing with complex arithmetic, to add or subtract two complex numbers, they need to be in rectangular form. It's trivial to see that \$(A + jB) + (C + jD) = (A + C) + j(B + D)\$. However, if you have two complex numbers in polar form (which is essentially what a phasor is), you need to convert the polar form into rectangular. Conversely, to multiply or divide complex numbers, it's most convenient to work in polar form.


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