Question

The locus of $z$ satisfying the inequality $\frac{z + 2i}{2z + i} < 1$ , where $z = x + iy$, is

• A. $x^2 + y^2 < 1$
• B. $x^2 - y^2 < 1$
• C. $x^2 + y^2 > 1$
• D. $2x^2 + 3y^2 < 1$

Answer 1

Answer: (C)

$x^2 + y^2 > 1$

Answer 2

Let $z = x + iy$
Given, $\frac{z + 2i}{2z + i} < 1$
$\Rightarrow \, \, \, \frac{\overline{x^2 + y + 2^2}}{\overline{ 2x^2 + 2y + 1^2}} < 1$
$\Rightarrow \, \, x^2 + y^2 + 4 + 4y < 4x^2 + 4y^2 + 1 + 4y$
$\Rightarrow \, \, 3x^2 + 3y^2 > 3$
$\Rightarrow \, \, x^2 + y^2 > 1$