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Complex Numbers and Quadratic Equations — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsComplex Numbers and Quadratic Equations1 Mark
MCQ
Choose the correct answer. The complex number $z$ which satisfies the condition $\Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|=1$ lies on:
  • A
    Circle $x^2 + y^2 = 1$
  • ✓
    The $x-$axis
  • C
    The $y-$axis
  • D
    The line $x + y = 1$
✓

Answer

Correct option: B.
The $x-$axis
Given that, $\Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|=1$
Let $\text{z}=\text{x}+\text{yi}$
$\therefore\ \Big|\frac{\text{i}+\text{x}+\text{yi}}{\text{i}-\text{x}-\text{yi}}\Big|=1$
$\Rightarrow\ \bigg|\frac{\text{x}-(\text{y}+1)\text{i}}{-\text{x}-(\text{y}-1)\text{i}}\bigg|=1$
$\Rightarrow|\text{x}+(\text{y}+1)\text{i}|=|-\text{x}-(\text{y}-1)\text{i}|$
$\Rightarrow\sqrt{\text{x}^2+(\text{y}+1)^2}=\sqrt{\text{x}^2+(\text{y}-1)^2}$
$\Rightarrow\ \text{x}^2+(\text{y}^2+1)^2=\text{x}^2+(\text{y}-1)^2$
$\Rightarrow(\text{y}+1)^2=(\text{y}-1)^2$
$\Rightarrow\text{y}^2+2\text{y}+1=\text{y}^2-2\text{y}+1$
$\Rightarrow2\text{y}=-2\text{y}$
$\Rightarrow2\text{y}+2\text{y}=0$
$\Rightarrow4\text{y}=0$
$\Rightarrow\text{y}=0$
$\Rightarrow\text{x-axis}$

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