Question

If $\left|z-\frac{3}{2}\right|=2$ , then the greatest value of $\left|z\right|$ is

• A. 1
• B. 2
• C. 3
• D. 4
• E. 5

Answer 1

Answer: (C)

Given, $\left|z-\frac{3}{z}\right|=2$
$\because |z|=\left|\left(z-\frac{3}{z}\right)+\frac{3}{z}\right|$
$\Rightarrow |z|\le \left|z-\frac{3}{z}\right|+\frac{3}{|z|}[\because |a+b|\le |a|+|b|]$
$\Rightarrow |z|\le 2+\frac{3}{|z|}$
$\Rightarrow |z{|}^2\le 2|z|+3$
$\Rightarrow |z{|}^2-2|z|-3\le 0$
$\Rightarrow (|z|-3)(|z|+1)\le 0$
$\Rightarrow -1\le |z|\le 3$
Hence, the greatest value of $|z|$ is 3 .