So the reactance of the inductor is jωL while that of capacitor is 1/jωC. My question is that starting from the differential equations that describe the dynamic behavior of these circuit components, how in the world did a person actually realize that Oh! lets put a j which represents an imaginary number?

Why is it merely j and not -j or some other j value? And why is j not in numerator of both the reactances?

  • 1
    \$\begingroup\$ Look in any decent electronics textbook. \$\endgroup\$
    – Barry
    Jun 11, 2015 at 23:21
  • \$\begingroup\$ What happens if you multiply top & bottom by \$j\$? \$\endgroup\$ Jun 11, 2015 at 23:28
  • \$\begingroup\$ The j comes from the Fourier transform of the governing equation. Similarly, that's where the \$\omega\$ comes from, too. \$\endgroup\$ Jun 11, 2015 at 23:58
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    \$\begingroup\$ Because Euler. \$\endgroup\$ Jun 12, 2015 at 0:09
  • \$\begingroup\$ When you move from cos and sin to complex exponentials to make equations easier, you introduce the imaginary unit j. In the complex plane multiplying by j means rotating the phasor by 90 degrees ccw; dividing is a 90 degrees cw rotation. Hence the j in the reactances reflect the phase shift of voltage with respect to current in those two ideal cases. \$\endgroup\$ Jun 12, 2015 at 0:28

2 Answers 2


This is a bit of a dummy's guide: -

When you look at the current thru a resistor by applying a sinewave across it, at every point on the waveform the ratio of V to I = R.

Now consider what this looks like for an inductor (or a capacitor): -

enter image description here
(source: johnhearfield.com)

At no points on either of the two traces (within one cycle) is there a constant ratio between V and I. So, if "+j" was used to shift the current waveform for an inductor by +90 it would align voltage and current for an inductor and you would get a meaningful relationship in the time domain.

For a capacitor "-j" is used and 1/j = -j: -

enter image description here

As you can hopefully see +j is a fixed 90 degrees rotation anticlockwise from the A position corresponding to 0 degrees on the chart. Hopefully you can see that -j is a rotation of 270 degrees (or -90)


The use of imaginary numbers follows naturally when analyzing oscillating motion, and particularly when analyzing electrical circuits which incorporate reactive components, where voltage and current are not identical. As has been suggested, do a little research to find out why.

As an interesting minor issue, the question of why electrical engineers use j rather than the mathematician's i is a bit murky. On the one hand, when complex notation was introduced, i was already in use for current (almost certainly from the German Intensität). So what to do? j was chosen, probably for 2 reasons. First, of course, it was the next letter in the alphabet, and was not spoken for. A second possibility is discussed here http://www.johndcook.com/blog/2013/04/23/why-j-for-imaginary-unit/ which suggests that it relates to representing complex numbers in an ij plane, rather than xy.

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    \$\begingroup\$ I always thought that j was chosen because I was already in use for current. V=IR, I for DC current and i for the AC component. Therefore the next letter j was chosen for the imaginary unit. \$\endgroup\$
    – John D
    Jun 12, 2015 at 1:31
  • \$\begingroup\$ When can write a number as a magnitude with phase and then treat this as a "composite" number and do maths with it using trignometric ratios and arithmatic. However, once we write this as a complex exponential..... that is when it becomes a mysteriour alien thing. \$\endgroup\$
    – quantum231
    Jul 12, 2015 at 23:50

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