Question

If $(x+iy)^{1/3}=a+ib$, where $x$, $y$ , $a$ , $b \in R$ , then $\frac{x}{a}-\frac{y}{b}=$

• A. $a^{2}-b^{2}$
• B. $-2(a^{2}+b^{2})$
• C. $2(a^{2}-b^{2})$
• D. $a^{2}+b^{2}$

Answer 1

Answer: (B)

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