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If x + iy1/...

Question

If $(x + i y)^{1 / 3} = a + i b$ then prove that:
$\frac{x}{a} + \frac{y}{b} = 4 (a^{2} - b^{2})$

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Solution

Now,
$(x + i y)^{1 / 3} = a + i b$
Cubic both sides we get,
$(x + i y) = (a + i b)^{3}$
or, $x + i y = a^{3} - 3 a b^{2} + i (3 a^{2} b - b^{3})$
Comparing both the real and complex part both sides we get,
$x = a^{3} - 3 a b^{2}$ ......(1) and $y = 3 a^{2} b - b^{3}$ .......(2).
Now
$\frac{x}{a} + \frac{y}{b}$
$= (a^{2} - 3 b^{2}) + (3 a^{2} - b^{2})$
$= 4 (a^{2} - b^{2})$ .

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