If $\omega$ is a complex number such that $\omega$ describes the set of complex numbers such that $|\omega|=q \ne 1$, then the complex number $$z = \omega + \frac{1}{\omega}$$ describes a conic. The distance between the foci of the conic described by $z$ is?
I attempted to rewrite this in cartesian form but to no avail. How do i proceed?
Let $q \in \mathbb{R}$, $q \ne 1$, and let $\omega$ range over all $\omega \in \mathbb{C}$ such that $|\omega| = q$. Then we can write $$ \omega = q \cos \theta + iq \sin \theta $$ and $$ \frac{1}{\omega} = \frac{1}{q} \cos \theta - \frac{i}{q} \sin \theta $$ where $\theta \in \mathbb{R}$. Therefore $$ z = \omega + \frac{1}{\omega} = \left(q + \frac{1}{q}\right) \cos \theta + i\left(q - \frac{1}{q}\right) \sin \theta. $$ That is, \begin{equation} z = a \cos \theta + ib \sin \theta,\tag{1}\end{equation} where $a = q + \frac{1}{q}$ and $b = q - \frac{1}{q}$. Equation ($1$) is the equation of an ellipse with semimajor axis $a$ and semiminor axis $|b|$, whose foci are therefore separated by the distance $2 \sqrt{a^2 - b^2} = 4$. Not coincidentally, $4$ is also the length of the line segment generated by $z = \omega + \frac{1}{\omega}$ where $|\omega| = 1$.