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COMPLEX NUMBERS AND QUADRATIC EQUATIONS — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCOMPLEX NUMBERS AND QUADRATIC EQUATIONS1 Mark
MCQ
If $\text{z}=\text{a}+\text{ib}$ lies in third quadrant, then $\frac{\bar{\text{z}}}{\text{z}}$ also lies in third quadrant if:
  • A
    $\text{a}>\text{b}>0$
  • B
    $\text{a}<\text{b}<0$
  • ✓
    $\text{b}<\text{a}<0$
  • D
    $\text{b}>\text{a}>0$
✓

Answer

Correct option: C.
$\text{b}<\text{a}<0$
Since, $\text{z}=\text{a}+\text{ib}$ lies in third quadrant.
$\Rightarrow\text{a}<0$ and $\text{b}<0 \ ...(1)$
Now,
$\frac{\bar{\text{z}}}{\text{z}}=\frac{\overline{\text{a}+\text{ib}}}{\text{a}+\text{ib}}$
$=\frac{\text{a}-\text{ib}}{\text{a}+\text{ib}}$
$=\frac{\text{a}-\text{ib}}{\text{a}+\text{ib}}\times\frac{\text{a}-\text{ib}}{\text{a}-\text{ib}}$
$=\frac{\text{a}^2+\text{i}^2\text{b}^2-2\text{abi}}{\text{a}^2-\text{i}^2\text{b}^2}$
$=\frac{\text{a}^2-\text{b}^2-2\text{abi}}{\text{a}^2+\text{b}^2}$
Since, $\frac{\bar{\text{z}}}{\text{z}}$ also lies in third quadrant.
$\Rightarrow\text{a}^2-\text{b}^2<0$
$\Rightarrow(\text{a}-\text{b})(\text{a}+\text{b})<0$
$\Rightarrow\text{a}-\text{b}>0$ and $\text{a}+\text{b}<0$
$\Rightarrow\text{a}>\text{b} \ ...(2)$
From (1) and (2),
$\text{b}<\text{a}<0$

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