Question

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

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

Answer 1

The correct option is C $(-\frac{\pi }{4},0)$

$z_1=\frac{kz}{(k+1)}$ represents any point lying on the line joining origin and $z$.
Given,
$|\left[\frac{kz}{k+1}\right]+2i|>\sqrt{2}$
Hence, $\frac{kz}{k+1}$ will lie outside the circle $|z_1+2i|=\sqrt{2}$.


So, $z$ should lie in the shaded region.
Now $△OPA$ is a rightangled triangle
$O{P}^2=O{A}^2-A{P}^2=4-2=2\Rightarrow OP=AP\Rightarrow \angle POA=\frac{\pi }{4}$
$\therefore -\frac{\pi }{4}<\arg \left(z\right)<0$