I have an electromagnetic wave with the magnetic field

$$\mathbf{B}=B_0\sin\left(\omega\left(t-\frac{a_zz}{c}\right)\right) \ \hat{x}.$$

I'm asked to write this in complex form.

I know that all time-harmonic fields may be written as

$$\mathbf{E}(r,t)=\text{Re}\{\mathbf{E}_s(r,t)e^{j\omega t} \}. \tag 1$$

I'm not sure how to use (2) to my advantage, I'd appreciate if someone could show me how to apply it. Why can't I just use Eulers formula for the sine?

$$B_0\sin\left(\omega\left(t-\frac{a_zz}{c}\right)\right)=B_0\frac{e^{j\omega t}e^{-j\omega\frac{a_zz}{c}}-e^{-j\omega t}e^{j\omega\frac{a_zz}{c}}}{2j}.$$

This is what they do in the answer sheet:

$$\mathbf{B}(r,t)=\text{Re}\{B_0(-j)e^{-j\omega\frac{a_zz}{c}} e^{j\omega t}\} \\ \mathbf{B}(r)=-jB_0 e^{-j\omega\frac{a_zz}{c}} \ \hat{x}.$$

Quite lost as to what is happening here. Where is the phasor? Why do they get rid of $e^{j\omega t}?$


1 Answer 1


If you use Euler's formula to write $$B_0\exp j(\omega t - \omega a_z z/c) = B_0\cos (\omega t -\omega a_z z/c) + jB_0\sin(\omega t - \omega a_z z/c),$$ then your original expression is obtained by multiplying this by $-j$ and taking the real part.

Then, if all you are interested in is the spatial dependence ${\bf B}(r)$ (rather than ${\bf B}(r,t)$), then yes, you can separate this from the multiplicative sinusoidal temporal dependence and just drop the $\exp (j\omega t)$ part.

  • $\begingroup$ So, why can't I simply take the imaginary part instead? $\endgroup$
    – Parseval
    Dec 24, 2019 at 10:21
  • 1
    $\begingroup$ @Parseval No reason. Just convention. But your equation 1 and the solution you didn't understand refer to the real part of a complex exponential. $\endgroup$
    – ProfRob
    Dec 24, 2019 at 10:35
  • $\begingroup$ But if I choose the imaginary part I'd get only $$B_0\sin(wt-wa_zz/c),$$ my $-j$ falls out, so this can't be correct. Or I'm just missing something very fundamental. $\endgroup$
    – Parseval
    Dec 24, 2019 at 19:07
  • $\begingroup$ @Parseval isn't that what you are trying to get? It appears to be what you started with. By taking the imaginary part you have implicitly ignored a j. $\endgroup$
    – ProfRob
    Dec 24, 2019 at 21:04

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