I have just started to learn about AC network analysis and have some questions about "j" (or "i" on my calculator), the imaginary unit. My book doesn't go into a great deal about this, and jumps right into formulas and substitutions (more practical approach, not theoretical). So, what exactly does J represent?

I see that if I draw a complex-plane (y-axis being imaginary, x-axis being real), and draw a unit circle on it, a 90° angle is \$\sqrt{-1}\$, which is "j". I see that I can use this substitution in phasor form when, say, solving for the voltage across a capacitor when the current through it is known:

$$V = \frac{I}{j \omega C}$$

Can someone help me understand this?

To be honest, this question is pretty vague because I'm not even sure how to ask about what J is; it's that foreign to me. I would like a common-sense explanation (big-picture) of it's meaning and purpose in AC circuit analysis. I'm not necessarily looking for a rigorous mathematical explanation (although any necessary mathematical explanation is welcome).

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    \$\begingroup\$ Algebra is case-sensitive. J and j are different things. \$\endgroup\$
    – TRiG
    Jul 30, 2014 at 16:49
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    \$\begingroup\$ You might want to look at the questions under the complex-numbers tag on math.SE: math.stackexchange.com/questions/tagged/… \$\endgroup\$
    – The Photon
    Jul 30, 2014 at 19:07
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    \$\begingroup\$ Of course what you find on math.SE will leave open the really interesting question: Why are complex numbers useful in engineering? \$\endgroup\$
    – The Photon
    Jul 30, 2014 at 19:09
  • \$\begingroup\$ @The Photon: The answer is on Wikipedia: en.wikipedia.org/wiki/Phasor I can summarize it here, but given the dynamics of voting on SE sites, it would be "wasted bullets". \$\endgroup\$
    – Fizz
    Jan 13, 2015 at 18:03
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    \$\begingroup\$ @The Photon: I was actually unpleasantly surprised by the Wikipedia article on Phasor because it failed to outline in its lead (i.e. introductory paragraphs nutshell) the main selling point of Phasors. So, in the past 24hrs, I've significantly improved the lead of the Wikipedia article on Phasor. I think its lead makes it a lot more obvious now why Phasors are used. Perhaps we should continue this discussion on the talk page of that article. \$\endgroup\$
    – Fizz
    Jan 14, 2015 at 17:08

4 Answers 4


If you put a minus sign in front of the number "5" it becomes "-5".

Try and look at this differently. Try thinking that it rotates the number "5" (tied to the origin by a piece of string of length 5) through 180 degrees to become "-5"

OK so far? Negative signs are the same as rotating thru 180 degrees...

Why not extend this further to produce something you can "stick" in front of a positive number that rotates it thru 90 degrees - in EE this is usually called "j" and it acts to rotate a value (about the origin) thru 90 degrees counter-clock wise i.e. if you did it twice (j*j) you'd get 180 degrees ("-").

From this gem of knowledge you can therefore say j*j = -1 therefore, j = \$\sqrt{-1}\$

Just as a minus sign can rotate any positive value thru 180 degrees it can rotate any vector or phasor thru 180 degrees. The same applies to the j operator - it rotates any vector or phasor thru 90 degrees counter clockwise.

EDIT - forgot part of question: -

substituting j into the impedance of a capacitor. Remember the basic formula for a capacitor is Q=CV and therefore differentiating the variables we get: -

\$ I = \dfrac{dQ}{dt} = C\dfrac{dV}{dt}\$

This tells us that for a sinewave applied voltage across a capacitor, the current will also be a sinewave but differentiated into a cosine like this: -

enter image description here

If you tried to calculate the impedance (V/I) of a capacitor from the V-I relationship you'd get into trouble because when I passes thru zero, V is NOT zero so you get infinities. If on the other hand you apply a "j" to bring current in phase with voltage the math works out fine - current and voltage are aligned and impedance based on instantaneous values of V/I makes sense.

I'm aware that you are just starting out so I've tried to keep this both accurate and simple (maybe too simple for some?).

If you look at the inductor, the "j" can be applied to the voltage to align it with the current hence "j" is in the numerator for inductive reactance and j is in the denominator for capacitive reactance. There are subtleties lying around here that hopefully will make sense as you learn more - it's actually no coincidence that "j" appears to "follow" omega when it comes to impedances - my explanation doesn't cover that and neither does your question!

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    \$\begingroup\$ I found your answer to be very helpful, especially with your mention of using j to bring the waveforms in phase; this helped me understand its use because I remember that voltage leads current by 90* for pure inductance, and vice-versa for pure cap. Thanks! \$\endgroup\$
    – asdf
    Jul 31, 2014 at 5:08
  • \$\begingroup\$ @Andy aka, does the 'j' serve any other purpose apart from enabling division between V and I when I is zero? \$\endgroup\$
    – penguin99
    Jun 11, 2019 at 9:34
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    \$\begingroup\$ @noorav it serves other purposes such as in solving transfer functions in filters and control systems. In my example above I was just using it to shift a voltage waveform to align it with a current waveform. You may be aware of the field of complex numbers. \$\endgroup\$
    – Andy aka
    Jun 11, 2019 at 9:43

In pure maths we use \$ i \$ to represent the prime square root of \$-1\$.

The other square root of \$-1\$ being \$-i\$.

If you imagine a number line with real numbers placed horizontally. We can now add a second number line going vertically containing the imaginary numbers.

We have now created a system of complex where every point on the plane is represented by a real and imaginary part e.g. \$ 4 + 3i\$ represents a point that is 4 units along the real axis and 3 units up the imaginary axis.

Because a point in two dimensional space can now be represented as a single number, calculations involving 2-dimensional vectors are simplified.

In electronics, when considering systems supplied by a single frequency sine wave, we are taught initially to draw phasor diagrams. Then later to use complex numbers to get to deal with these problems.

We also use \$ j \$ instead of \$ i \$ but the meaning is identical. It’s just to avoid confusion because in electronics \$i\$ is often used for current.

If you would like a little more insight take a look at this question: What are imaginary numbers? from the Mathematics Stack Exchange site.

Or take a look here: A Visual, Intuitive Guide to Imaginary Numbers.

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    \$\begingroup\$ Thanks for your help and references to some additional reading! \$\endgroup\$
    – asdf
    Jul 31, 2014 at 5:09
  • \$\begingroup\$ Note: while complex numbers can be used to represent the same pair of numbers or the same point as an \$\mathbb{R}^2\$ vector does, the vector and complex number algebras are quite different in how they transform the coordinates once you go past addition and subtraction. There’s no vector algebra primitive operation equivalent of multiplying complex numbers, for example. \$\endgroup\$ Sep 2, 2022 at 13:17

In math someone asked the question:

What's the solution to x^2 = -1 ?

They invented a number and said let's call it "j".

They worked out the consequences of doing this. They found that it didn't lead to any contradictions within the realm of the existing mathematics.

Note that you might think, "ok, why not just introduce a letter every time you have something un-solvable? I'll just call 1/0 = f".

Try it. It doesn't always work because the existing rules of arithmetic break down. For example you can show that defining 1/0 = f allows you to show that 1=2, or 1=3, ...

So mathematically it works and didn't lead to any contradictions. Suddenly we have a way to "pack in" two pieces of information into a single number because of the way you can represent a complex number: on a real/imaginary plane. Suddenly we can manipulate a NUMBER that contains both magnitude and phase just the same way as we manipulate "regular numbers". This is quite useful.

In electronics it's quite convenient to be able to pack in two pieces of information into one number. So it's quite convenient to make use of complex numbers. That's all it is. We just so happen to want to keep track of both a magnitude and a phase -- this tool of mathematics which has in many ways just been invented out of thin air but doesn't break any rules allows us to do just that. So let's use it.

  • \$\begingroup\$ You're skipping over some pretty important details here. Imaginary numbers aren't just a way of combining two arbitrary real numbers into a vector; the structure of complex numbers makes operations on the real/imaginary pair behave in a specific way. \$\endgroup\$
    – user39382
    Jun 2, 2017 at 18:53
  • \$\begingroup\$ @duskwuff: I think his point was that once one decides that j represents one of the two roots of x^2=-1, there's no need to invent a structure for them, since the structure of complex numbers [e.g. multiplying (a+bj) by (c+dj) will yield (ac-bd)+(ad+bc)j] follows from the combining the laws of arithmetic with that one additional axiom. \$\endgroup\$
    – supercat
    Jun 2, 2017 at 20:25
  • \$\begingroup\$ @supercat Right. What I'm trying to get at is that there's some physical significance to that structure -- it's not just some random made-up mathematical trick. \$\endgroup\$
    – user39382
    Jun 2, 2017 at 20:31

In mathematics imaginary unit is a very helpful number used to solve equations with higher than 2 order. It was introduced just.... to the test, and it works pretty until today. This provides for obtaining at least one root in every polynomial.

In electronics imaginary unit represents the energy stored in our circuit. So, in capacitor, it is the energy stored in it. It also represents phase shift in circuit, when we are dealing with sinusoidal signals.

I think you should more precise your question, or just write questions which bother you in points.

For example... If your circuit's impedance will be represented only by imaginary unit, not by real, your bill for energy will be... zero :)


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