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Algebra of Complex Numbers

If √x+iy=a+...

Question

If $\sqrt[3]{x + i y} = a + i b$ then prove that $\frac{x}{a} + \frac{y}{b} = 4 (a^{2} - b^{2})$ .

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Solution

Given,
$\sqrt[3]{x + i y} = a + i b$
Cubing 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 sides we get,
$x = a^{3} - 3 a b^{2}$ and $y = 3 a^{2} b - b^{3}$ .
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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then prove that:

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, then show that

\frac{x}{a} + \frac{y}{b} = 4 (a^{2} - b^{2})

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