Finding solutions for Real and Imaginary parts of a Complex Number.

Question

This is my starting equation for complex number, $ \hat{k} $

$$ \hat{k}^2 = \mu \epsilon \omega - \mu \sigma \omega^2 i $$

I assumed $ \hat{k} = \alpha + \beta i $ where $ \alpha , \beta \in \mathbb{R} $

I then derived the following:

$$ \alpha^4 - \mu \epsilon \omega \alpha^2 - \frac{1}{4} \mu^2 \sigma^2 \omega^2 = 0 $$

Which gives us four solutions for $ \alpha $.

If we consider $\alpha ^2 = x $ and solve for $ x $, we get 2 solutions. Griffith's Intro to Electrodynamics (3rd ed.) pg 394, states that these two are only two unique solutions for $ \alpha $ and $ \beta $. And it seems to be the solutions only used the degree 4 equation I am solving above...

But that doesn't make sense,

1. how can we get the solution for $ \beta $ from the equation solving for $ \alpha $?
2. Shouldn't we also consider that there are a total of 8 eight solutions, 4 for $ \alpha $ and $ \beta $ each?

How can we then find unique solutions for $ \alpha $ and $ \beta $ ? Does it have to do with the fact that $ \alpha , \beta \in \mathbb{R} $ ?

Answer 1

The initial equation, namely $$ k^2 = (\alpha + i\beta)^2 = \alpha^2 - \beta^2 + 2i\alpha\beta = \mu\epsilon\omega - i\mu\sigma\omega^2 $$ is equivalent to the system $$ \left\{ \begin{align} \alpha^2 - \beta^2 &= \mu\epsilon\omega & (1) \\ 2\alpha\beta &= -\mu\sigma\omega^2 & (2) \end{align} \right. $$ when separating the real and imaginary parts. The second equation permits to make the substitution $$ \beta = -\frac{\mu\sigma\omega^2}{2\alpha} \quad\quad (3) $$ into the first one, hence the biquadratic equation you wrote, i.e. $$ \alpha^4 - \mu\epsilon\omega\alpha^2 - \frac{1}{4}\mu^2\sigma^2\omega^\color{red}{4} = 0. $$ So the answer to your first question is simply given by Eq. (3). Now, the above equation is solved (after the algebraic clean-up) by $$ \alpha_\pm^2 = \frac{1}{2}\mu\epsilon\omega \left(1 \pm \sqrt{1 + \left(\frac{\sigma}{\epsilon\omega}\right)^2}\right), $$ hence $\alpha_-^2 < 0$, which is impossible, since it should be a (real) square. In the same spirit, you can exclude another solution when computing $\alpha$ by imposing it to be positive.

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Addendum. An alternative parametrization of $\alpha$ and $\beta$ is represented by the following hyperbolic polar coordinates : $$ \begin{cases} \alpha = \rho\cosh\theta \\ \beta = \rho\sinh\theta \end{cases} $$ In consequence, one has : $$ \begin{cases} \rho^2 = \mu\epsilon\omega \\ \rho^2\sinh2\theta = -\mu\sigma\omega^2 \end{cases} $$ This system can be solved straightforwardly and possesses the advantage to avoid to generate extraneous solutions $-$ due to the fact that the hyperbolic sinus is bijective and $\rho$ can't be negative obviously, because of its role of radius.