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)

${\left(x+iy\right)}^{1/3}=a+ib$
$\Rightarrow x+iy={\left(a+ib\right)}^3$
$\Rightarrow x+iy$
$=a^3-i{b}^3+i3a^{2}b-3a{b}^2$
$=a^3-3a{b}^2+i\left(3a^{2}b-b^3\right)$
$\Rightarrow x=a^3-3a{b}^2$ and $y=3a^{2}b-b^3$
So, $\frac{x}{a}-\frac{y}{b}$
$=a^2-3b^2-3a^2+b^2$
$=-2a^2-2b^2$
$=-2\left(a^2+b^2\right)$