Question

if $\omega$ is complex number such that $|$ $\omega$ $|$ $\ne 1$ then the complex number z = $\omega$ + $\frac{1}{\omega }$ describes

• A. ellipse
• B. circle
• C. parabola
• D. straight line

Answer 1

The correct option is A

ellipse

We will take z = x + iy and try to establish some relation between x and y. To do this we have to find the real part and imaginary part of $\omega$ + $\frac{1}{\omega }$. One way to proceed is by taking w = h + ik. If we proceed in that way, it may be difficult to multiply various terms and then simplify. So we will look for other alternatives. If we take $\omega$ = $r{e}^{i\theta }$ and $\frac{1}{\omega }$ will be easy to calculate and there won't be terms also.
z = $\omega$ + $\frac{1}{\omega }$
$\Rightarrow$ x + iy = $r{e}^{i\theta }$ + $\frac{1}{r}$$e^{-i\theta }$ = r(cos $\theta$ + 1 sin$\theta$) + $\frac{1}{r}$(cos $\theta$ - sin$\theta$)
x + iy = $(r+\frac{1}{r})$cos $\theta$+i $(r-\frac{1}{r})$sin$\theta$
$\Rightarrow x=(r+\frac{1}{r})cos\theta ,y=(r-\frac{1}{r})sin\theta$
We got x and y in terms of other variables. We will try to eliminate $\theta$.
$\left(\frac{x}{r+\frac{1}{r}}\right)$ = cos $\theta$, $\left(\frac{y}{r-\frac{1}{r}}\right)$ = sin $\theta$
We know $co{s}^2$$\theta$ + $si{n}^2$$\theta$ = 1
$\Rightarrow$($\frac{x^2}{(r+\frac{1}{r}{)}^2}$) + ($\frac{y^2}{(r-\frac{1}{r}{)}^2}$) = 1, which is an ellipse.