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

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCOMPLEX NUMBERS AND QUADRATIC EQUATIONS4 Marks
Question
If $(x + iy)^3 = u + iv$, then show that $\frac{\text{u}}{\text{x}}+\frac{\text{v}}{\text{y}}=4(\text{x}^2-\text{y}^2).$
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Answer

$(\text{x}+\text{iy})^3=\text{u}+\text{iv}$ $\Rightarrow\text{x}^3+(\text{iy})^3+3.\text{x}.\text{iy}(\text{x}+\text{iy})=\text{u}+\text{iv}$ $\Rightarrow\text{x}^3+\text{i}^3\text{y}^3+3\text{x}^2\text{yi}+3\text{x}\text{y}^2\text{i}^2=\text{u}+\text{iv}$ $\Rightarrow\text{x}^3-\text{i}\text{y}^3+3\text{x}^2\text{yi}-3\text{x}\text{y}^2=\text{u}+\text{iv}$ $\Rightarrow(\text{x}^3-3\text{x}\text{y}^2)+\text{i}(3\text{x}^2\text{y}-\text{y}^3)=\text{u}+\text{iv}$ On equating real and imaginary parts, we obtain $\text{u}=\text{x}^3-3\text{x}\text{y}^2,\ \text{v}=3\text{x}^2\text{y}-\text{y}^3$ $\therefore\ \frac{\text{u}}{\text{x}}+\frac{\text{v}}{\text{y}}=\frac{\text{x}^3-3\text{x}\text{y}^2}{\text{x}}+\frac{3\text{x}^2\text{y}-{\text{y}^3}}{\text{y}}$ $=\frac{\text{x}(\text{x}^2-3\text{y}^2)}{\text{x}}+\frac{\text{y}(3\text{x}^2-\text{y}^2)}{\text{y}}$ $=\text{x}^2-3\text{y}^2+3\text{x}^2-\text{y}^2$ $=4\text{x}^2-4\text{y}^2$ $=4(\text{x}^2-\text{y}^2)$ $\therefore\ \frac{\text{u}}{\text{x}}+\frac{\text{v}}{\text{y}}=4(\text{x}^2-\text{y}^2)$ Hence, proved.

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