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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
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where $k$ is equal to
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The correct answer is:
4
$z^{1 / 3}=a-i b \Rightarrow z=(a-i b)^{3}$
$\therefore x+i y=a^{3}+i b^{3}-3 i a^{2} b-3 a b^{2}$
$\Rightarrow x=a^{3}-3 a b^{2} \Rightarrow \frac{x}{a}=a^{2}-3 b^{2}$
and $y=b^{3}-3 a^{2} b \Rightarrow \frac{y}{b}=b^{2}-3 a^{2}$
So, $\frac{x}{a}-\frac{y}{b}=4\left(a^{2}-b^{2}\right)$
$\therefore x+i y=a^{3}+i b^{3}-3 i a^{2} b-3 a b^{2}$
$\Rightarrow x=a^{3}-3 a b^{2} \Rightarrow \frac{x}{a}=a^{2}-3 b^{2}$
and $y=b^{3}-3 a^{2} b \Rightarrow \frac{y}{b}=b^{2}-3 a^{2}$
So, $\frac{x}{a}-\frac{y}{b}=4\left(a^{2}-b^{2}\right)$
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