Question:
The locus of $z$ satisfying the inequality $\frac{z + 2i}{2z + i} < 1$ , where $z = x + iy$, is
The locus of $z$ satisfying the inequality $\frac{z + 2i}{2z + i} < 1$ , where $z = x + iy$, is
Updated On: Jun 17, 2022
- $x^2 + y^2 < 1 $
- $x^2 - y^2 < 1 $
- $x^2 + y^2 > 1 $
- $2x^2 + 3y^2 < 1 $
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The Correct Option is C
Solution and Explanation
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 $
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 $
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Concepts Used:
Complex Number
A Complex Number is written in the form
a + ib
where,
- “a” is a real number
- “b” is an imaginary number
The Complex Number consists of a symbol “i” which satisfies the condition i^2 = −1. Complex Numbers are mentioned as the extension of one-dimensional number lines. In a complex plane, a Complex Number indicated as a + bi is usually represented in the form of the point (a, b). We have to pay attention that a Complex Number with absolutely no real part, such as – i, -5i, etc, is called purely imaginary. Also, a Complex Number with perfectly no imaginary part is known as a real number.