Q. The locus of $z$ satisying the inequality $\left| \frac{z + 2i}{2z + i} \right| < 1$ where $z = x + iy$, is
A) $x^{2} + y^{2} < 1$
B) $x^{2} - y^{2} < 1$
C) $x^{2} + y^{2} > 1$
D) $2x^{2} + 3y^{2} < 1$
Explanation
(c)
Let $z = x + iy$
Given, $\left| \frac{z + 2i}{2z + i} \right| < 1$
$\Rightarrow \ \ \ \ \ \frac{\sqrt{(x)^{2} + (y + 2)^{2}}}{(2x)^{2} + (2y + 1)^{2}} < 1$
$\Rightarrow \ \ \ \ x^{2} + y^{2} + 4 + 4y < 4x^{2} + 4y^{2} + 1 + 4y$
$\Rightarrow \ \ \ 3x^{2} + 3y^{2} > 3$
$\Rightarrow \ \ \ x^{2} + y^{2} > 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.