Question

I. lf $|z-\frac{2}{z}|=2$ then the greatest value of $|z|$ is $\sqrt{3}+1$
II. $|z+1|-|z-1|=\frac{3}{2}$ then the least value of $\left|z\right|=\frac{3}{4}$

• A. Only I is true
• B. Only II is true
• C. both I and II are true
• D. Neither I nor II are true

Answer 1

The correct option is C both I and II are true

I. We have $|z|=|z+\frac{2}{z}-\frac{2}{z}|$ $\le |z|+\frac{2}{z}+\left(\frac{-2}{z}\right)|$
$\le |z+\frac{2}{z}|+|-\frac{2}{z}|$
$\therefore |z|\le 2+\frac{2}{|z|}$
$\therefore |z{|}^2-2|z|-2\le 0$
$\therefore (|z|-1+\sqrt{3})(|z|-1-\sqrt{3})\le 0$
$1-\sqrt{3}\le |z|\le 1+\sqrt{3}$
Hence max value of z is $1+\sqrt{3}$

II. $(1+x)-(1-x)=\frac{3}{2}$
$2x=\frac{3}{2}$
$x=\frac{3}{4}$