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  4. If z = x + iy, z1/3 = a - ib, then (x/a) - (y/b) = k (a2 - b2) where k is equal to

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

BITSATBITSAT 2014

Solution:

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