First slide

Algebra of complex numbers

Question

If $(x + i y)^{1 / 3} = a + i b,$ then $\frac{x}{a} + \frac{y}{b}$ equals

Easy

A

$4 (a^{2} - b^{2})$
B

$2 (a^{2} - b^{2})$
C

$2 (a^{2} + b^{2})$
D

none of these

Solution

$(x + i y)^{1 / 3} = a + i b$
$\begin{array}{l} \Rightarrow x + i y = (a + i b)^{3} = a^{3} + 3 a^{2} (i b) + 3 a (i b)^{2} + (i b)^{3} \\ \Rightarrow x + i y = (a^{3} - 3 a b^{2}) + (3 a^{2} b - b^{3}) i \end{array}$
Equating real and imaginary parts, we get
$x = a^{3} - 3 a b^{2}$ and $y = 3 a^{2} b - b^{3}$
$\Rightarrow \frac{x}{a} = a^{2} - 3 b^{2}$ and $\frac{x}{b} = 3 a^{2} - b^{2} .$
Thus, $\frac{x}{a} + \frac{y}{b} = 4 a^{2} - 4 b^{2} = 4 (a^{2} - b^{2})$

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