Question

If $z$ is a complex number lying in the fourth quadrant of Argand plane and $|\frac{kz}{(k+1)}+2i|>\sqrt{2}$ for all real value of $k(k\ne -1)$, then range of arg(z) is

• A. $(-\frac{\pi }{8},0)$
• B. $(-\frac{\pi }{6},0)$
• C. $(-\frac{\pi }{4},0)$
• D. none of these

Answer 1

Answer: (C)

$(-\frac{\pi }{4},0)$

$|\frac{kz}{k+1}+2i|>\sqrt{2}\Rightarrow |z+\frac{2i(k+1)}{k}|>\frac{\sqrt{2}(k+1)}{k}$
From the fig:
In right angled $△$, $ACB:\angle ACB=\arcsin \frac{BC}{AB}=\frac{\frac{\sqrt{2}(k+1)}{k}}{\frac{2(k+1)}{k}}=\frac{\pi }{4}$
$\therefore arg\left(z\right)\in (-\frac{\pi }{4},0)$
Hence, option 'C' is correct.