Question

Let $z$ be a complex number such that $|z+\frac{1}{z}|=2$.

If $\left|z\right|=r_1$ and $r_2$ for $\arg z=\frac{\pi }{4}$ then

$|r_1-r_2|=$

• A. $\frac{1}{\sqrt{2}}$
• B. $1$
• C. $\sqrt{2}$
• D. $2$

Answer 1

Answer: (C)

$\sqrt{2}$

${|z+\frac{1}{z}|}^2=r^2+\frac{1}{r^2}+2r.\frac{1}{r}\cos (\theta +\theta )=4$
or $r^2+\frac{1}{r^2}=4-2\cos 2\theta$
$\therefore {(r-\frac{1}{r})}^2=2(1-\cos 2\theta )=4\sin {}^2\theta$
$|r_1-r_2|=|r-\frac{1}{r}|=2\sin \theta =2\sin \frac{\pi }{4}=\sqrt{2}$