Question

If $\omega$ is a complex number such that $\left|\omega \right|=r\ne 1$ then $z=\omega +\frac{1}{\omega }$ describes a conic. The distance between the foci is:

• A. $2$
• B. $2(\sqrt{2}-1)$
• C. $3$
• D. $4$

Answer 1

Answer: (D)

$4$

Ans. $\left(d\right)$.
If $\omega =r{e}^{i\theta }$ then $\frac{1}{\omega }=\frac{1}{r}{e}^{-i\theta }$
$\therefore z=(\omega +\frac{1}{\omega })$
$=r(\cos \theta +i\sin \theta )+\frac{1}{r}(\cos \theta -i\sin \theta )$
$\therefore x=(r+\frac{1}{r})\cos \theta$, $y=(r-\frac{1}{r})\sin \theta$
Eliminating $\theta$, $\frac{x^2}{{(r+\frac{1}{r})}^2}+\frac{y^2}{{(r-\frac{1}{r})}^2}=1$
Above represents an ellipse and distance between foci is
$2ae=2\sqrt{a^2(1-\frac{b^2}{a^2})}=2\sqrt{a^2-b^2}=2\sqrt{4}=4$