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

Answer 1

Answer: (C)

3

Answer 2

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