Question

If $\left|w\right|=2$ then the set of points $z=w-\left(1/w\right)$ is contained in or equal to

• A. An ellipse with eccentricity $\frac{3}{5}$
• B. An ellipse with eccentricity $\frac{3}{7}$
• C. An ellipse with eccentricity $\frac{4}{5}$
• D. An ellipse with eccentricity $\frac{4}{7}$

Answer 1

The correct option is C An ellipse with eccentricity $\frac{4}{5}$

Given $|w|=2⟹w\overline{w}=|w{|}^2=4$

It follows that $z=w-\left(1/w\right)=w-\left[\frac{\overline{w}}{4}\right].$

Let $z=x+iy$ and $w=u+iv$ with $x,y,u,v\in R,$ then the above can be written as:

$x+iy=u+iv-\frac{1}{4}(u-iv)=\frac{3}{4}u+i\frac{5}{4}v⟹x=\frac{3}{4}u$ and,$y=\frac{5}{4}v$

The condition $|w{|}^2=4⟺u^2+v^2=4$ then gives the locus of $z$ as the ellipse:

$(\frac{4}{3}x{)}^2+(\frac{4}{5}y{)}^2=4⟺\frac{x^2}{9}+\frac{y^2}{25}=\frac{1}{4}$

Hence $e=\sqrt{1-\frac{a^2}{b^2}}=\frac{4}{5}$......[for $b>a$]