Q47 / 125 · BITSAT 2014 · Mathematics

If $z=x+i y, z^{1 / 3}=a-i b,$ then $\frac{x}{a}-\frac{y}{b}=k\left(a^{2}-b^{2}\right)$ where $k$ is equal to

  • A
    1
  • B
    2
  • C
    3
  • Correct answer
    D
    4

Solution

$z^{1 / 3}=a-i b \Rightarrow z=(a-i b)^{3}$ $\therefore x+i y=a^{3}+i b^{3}-3 i a^{2} b-3 a b^{2}$ $\Rightarrow x=a^{3}-3 a b^{2} \Rightarrow \frac{x}{a}=a^{2}-3 b^{2}$ and $y=b^{3}-3 a^{2} b \Rightarrow \frac{y}{b}=b^{2}-3 a^{2}$ So, $\frac{x}{a}-\frac{y}{b}=4\left(a^{2}-b^{2}\right)$

Hence, the correct answer is option: 'D'.

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