Solving two linear equations with complex coefficeints

Question

With complex coefficients $\alpha$ and $\beta$ and a real parameter $\omega$, how can we find the value(s) of $\omega$ such the following equations are satisfied

\begin{equation} | \alpha \cos \omega + \beta \sin \omega |^2 =1/2 \end{equation}, and

\begin{equation} | \beta \sin \omega - \alpha \cos \omega |^2 =1/2? \end{equation}

Edit: There was a correction. The question is perfect now.

Answer 1

Let $\alpha = \alpha_1 + i\alpha_2$ and $\beta = \beta_1 + i\beta_2$, $\alpha_1, \alpha_2, \beta_1, \beta_2 \in \mathbb{R}$.

The equation is , $$ |\alpha\cos{\omega} + \beta\sin{\omega}|^2 = \frac{1}{2}. $$

By substituting $\alpha = \alpha_1 + i\alpha_2$ and $\beta = \beta_1 + i\beta_2$ into this equation, one has

$$ |(\alpha_1 + i\alpha_2)\cos{\omega} + (\beta_1 + i\beta_2)\sin{\omega}|^2 = \\ |(\alpha_1\cos{\omega} + \beta_1\sin{\omega}) + i(\alpha_2\cos{\omega} + \beta_2\sin{\omega})|^2 = \\ (\alpha_1\cos{\omega} + \beta_1\sin{\omega})^2 + (\alpha_2\cos{\omega} + \beta_2\sin{\omega})^2 = \frac{1}{2} \Leftrightarrow \\ (\alpha^2_1+\alpha^2_2)\cos^2{\omega} + 2(\alpha_1\beta_1 + \alpha_2\beta_2)\cos{\omega}\sin{\omega} + (\beta^2_1 + \beta^2_2)\sin^2{\omega} = \frac{1}{2}(cos^2{\omega} + \sin^2{\omega}) \Leftrightarrow \\ (\alpha^2_1+\alpha^2_2-\frac{1}{2})\cos^2{\omega} + 2(\alpha_1\beta_1 + \alpha_2\beta_2)\cos{\omega}\sin{\omega} + (\beta^2_1 + \beta^2_2-\frac{1}{2})\sin^2{\omega} = 0. $$

Now, let $A = \alpha^2_1+\alpha^2_2-\frac{1}{2}$, $B = 2(\alpha_1\beta_1 + \alpha_2\beta_2)$, $C = \beta^2_1 + \beta^2_2-\frac{1}{2}$, and let us us divide both sides of the equation by $\sin^2{\omega}$ assuming that $\sin{\omega}\ne 0$, i.e., $\omega\ne \pi n, n = 0, \pm 1, \pm 2, \ldots$.

Then we have a new equation

$$ A\cot^2{\omega} + B\cot{\omega} + C = 0 \Leftrightarrow |u = \cot{\omega}| \Leftrightarrow Au^2 + Bu + C = 0, $$

which is a quadratic equation.

Answer 2

We can use the $|z|^2=z\bar{z}$

The two relations become:

$(\alpha \cos \omega+\beta \sin\omega)(\overline{\alpha} \cos \omega+\overline{\beta} \sin\omega)=\frac12$

and $(\beta \sin\omega-\alpha \cos \omega)(\overline{\beta} \sin\omega-\overline{\alpha} \cos \omega)=\frac12$

Let $C=\cos \omega$ and $S=\sin\omega$

We have by expanding

$|\alpha|^2 C^2+|\beta|^2 S^2+\alpha\overline{\beta}CS+\overline{\alpha}\beta CS=\frac12$

$|\alpha|^2 C^2+|\beta|^2 S^2-\alpha\overline{\beta}CS-\overline{\alpha}\beta CS=\frac12$

Hence, by difference and sum,

$(\alpha \overline{\beta}+\overline{\alpha}\beta)CS=0$

and $2(|\alpha|^2C^2+|\beta|^2S^2)=1$

There remains a distinction between cases

$\alpha \overline{\beta}+\overline{\alpha}\beta=0$ or $\alpha \overline{\beta}+\overline{\alpha}\beta\neq 0$

...