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Complex Numbers — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSComplex Numbers1 Mark
MCQ
If $(\text{x}+\text{iy})^{\frac{1}{3}}=\text{a}+\text{ib,}$ then $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=$
  • A
    0
  • B
    1
  • C
    -1
  • ✓
    none of these
✓

Answer

Correct option: D.
none of these
$(\text{x}+\text{iy)}^{\frac{1}{3}}=\text{a}+\text{ib}$
Cubing on both the sides, we get:
$\text{x}+\text{iy}=(\text{a}+\text{ib})^3$
$\Rightarrow\text{x}+\text{iy}=\text{a}^3+(\text{ib})^3+3\text{a}^2\text{bi}+3\text{a}(\text{bi})^2$
$\Rightarrow\text{x}+\text{iy}=\text{a}^3+\text{i}^3\text{b}^3+3\text{a}^2\text{bi}+3\text{i}^2\text{ab}^2$
$\Rightarrow\text{x}+\text{iy}=\text{a}^3-\text{i}\text{b}^3+3\text{a}^2\text{bi}-3\text{ab}^2 \ (\because\text{i}^2=-1,\text{i}^3=-\text{i})$
$\Rightarrow\text{x}+\text{iy}=\text{a}^3-3\text{a}\text{b}^2+\text{i}(-\text{b}^3+3\text{a}^2\text{b})$
$\therefore\text{x}=\text{a}^3-3\text{a}\text{b}^2$ and $\text{y}=3\text{a}^2\text{b}-\text{b}^3$
or, $\frac{\text{x}}{\text{a}}=\text{a}^2-3\text{b}^2$ and $\frac{\text{y}}{\text{b}}=3\text{a}^2-\text{b}^2$
$\Rightarrow\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=\text{a}^2-3\text{b}^2+3\text{a}^2-\text{b}^2$
$\Rightarrow\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=4\text{a}^2-4\text{b}^2$

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