I'm attempting to code an AC power circuit simulator (for fun, imagine that) and I'm running into a mathematical stump. I've simplified the problem into these equations: $$a=bx+c$$ $$d=ex+f$$ $$P=|a||d|\cos(\theta_a-\theta_d)=a_R d_R+a_I d_I(=a\cdot d)$$ Where $a, b, c, d, e, f,$ and $x$ are all complex numbers, $P$ is a real number, and we know the values of $|a|, b, c, e, f,$ and $P$.

I know you can't really do a dot product between complex numbers but that's kind of how I'm thinking of this last equation (most of these complex numbers are actually phasors, or vector-wannabes anyways).

I need to find a way to express $x$ in terms of the known values. Is this even possible without knowing $\theta_a$ or $d$?

I don't seem to be getting anywhere by breaking down the first two equations into Real and Imaginary parts - it seems to be making more unknown values, which just makes this messier. But if I commit to polar form I'm finding it hard to deal with the cosine. Thank you for taking a look.

  • $\begingroup$ The dot product between complex numbers is defined : What is the dot product of complex vectors? $\endgroup$
    – Conan
    Commented Jul 28, 2022 at 3:34
  • $\begingroup$ @Conan That entry refers to the dot product of complex vectors, a vector with complex coordinates, while here we're dealing with complex scalars. The dot product of a complex vector would have us multiply a complex scalar in one vector by the conjugate of the similarly-indexed complex scalar in the other vector - the way I'm "using" a dot product here is different. $\endgroup$
    – ABradford
    Commented Jul 28, 2022 at 11:00
  • $\begingroup$ Welcome! That's a great first question for someone new to this site. $\endgroup$
    – Toby Mak
    Commented Jul 29, 2022 at 2:25
  • $\begingroup$ @TobyMak Thank you! $\endgroup$
    – ABradford
    Commented Jul 30, 2022 at 3:15

1 Answer 1


Let $a = |a|e^{i\theta}$ (your $\theta_a$, but I'm dropping the subscript as a needless pain). The first equation tells us that $$x = \frac{|a|e^{i\theta} - c}b$$ so if we can figure out $\theta$, then we can figure out $x$.

The second equation tells us that $$d = \frac{e|a|}be^{i\theta} +\left(f-\frac cb\right)$$

Now, $a\cdot d = \Re(\bar ad)$ ($\Re(z)$ denotes the real part of $z$). So the third equation tells us that $$\Re\left[|a|e^{-i\theta}\left(\frac{e|a|}be^{i\theta} + f-\frac cb\right)\right] = P\\\Re\left(\frac{e|a|}b\right) + \Re\left[\left(f-\frac cb\right)e^{-i\theta}\right] = \frac P{|a|}$$

Letting $Q = \frac P{|a|} - \Re\left(\frac{e|a|}b\right)$ and $re^{i\phi} = f-\frac cb$, we get $$\Re\left(re^{i(\phi - \theta)}\right) = Q$$ $$\cos(\phi - \theta) = \frac Qr$$ As $Q, r, \phi$ are all defined by known quantities, you can solve the equation for $\theta$ (in general, there will be two non-equivalent solutions), and obtain $x$ using the equation for it above.


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