constrained complex number equation requiring imaginary part to be zero

Question

Is there an easy way to solve the following equation for $k_1$ and $k_2$ under the constraint $|k_1|=|k_2|=1$?

$\operatorname{Im}(k_1b_1 - jk_2b_2) = \operatorname{Im}(k_2b_2 - jk_1b_1) =0$

where $b_1$ and $b_2$ are arbitrary complex numbers (but neither are zero)?

Is $j$ the square root of $-1$ here? – Rory Daulton Jan 03 '16 at 22:15 — Rory Daulton, Jan 03 '16 at 22:15
It will help to write the numbers in polar form – Ben Grossmann Jan 03 '16 at 22:25 — Ben Grossmann, Jan 03 '16 at 22:25

score 1 · Accepted Answer · answered Jan 03 '16 at 22:34

Saying that $\operatorname{Im}(z)=0$ is the same as saying $z=\bar{z}$ (the conjugate). So you have $$ k_1b_1-jk_2b_2= \overline{k_1}\,\overline{b_1}+j\overline{k_2}\,\overline{b_2} $$ Similarly, $$ k_2b_2-jk_1b_1= \overline{k_2}\,\overline{b_2}+j\overline{k_1}\,\overline{b_1} $$ Multiply the second equation by $j$ to get $$ k_1b_1+jk_2b_2= -\overline{k_1}\,\overline{b_1}+j\overline{k_2}\,\overline{b_2} $$ Sum the two relations and you have $$ 2k_1b_1=2j\overline{k_2}\,\overline{b_2} $$ Now you can have $k_1$ in terms of $k_2$ and substitute in the first equation.

Finally, the hypothesis that $|k_1|=1$ tells you $\overline{k_1}=k_1^{-1}$ and similarly for $k_2$.

That's interesting; that tells me that my initial constraint of $|k_1| = |k_2| = 1$ can't always be met. (e.g. if $b_1 = 2+3j$ and $b_2 = 4+5j$); also that if I have one solution $(k_1,k_2)$ then $(k_1e^{j\theta}, k_2e^{-j\theta})$ is also a solution.for any arbitrary $\theta$. — Jason S, Jan 04 '16 at 00:15

constrained complex number equation requiring imaginary part to be zero

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