Question

If $z=x-iy$ and $z^{\frac{1}{3}}=a+ib$, then $\frac{\left(\frac{x}{a}+\frac{y}{b}\right)}{a^2+b^2}=$

• A. -2
• B. -1
• C. 1
• D. 2

Answer 1

Answer: (A)

Given,
$z=x-iy$ and $z^{\frac{1}{3}}=a+ib$
$z^{1/3}=a+ib$
Take cube on both sides, we get
${\left(z^{1/3}\right)}^3=(a+ib{)}^3$
$\Rightarrow z=a^3+(ib{)}^3+3a^{2}ib+3a(ib{)}^2$
$\Rightarrow z=a^3+i^3{b}^3+3a^{2}ib+3a{b}^2{i}^2$
$\Rightarrow z=a^3-i{b}^3+3a^{2}bi-3a{b}^2$
$\left[\because i^2=-1\right]$
$\Rightarrow z=\left(a^3-3a{b}^2\right)-i\left(b^3-3a^{2}b\right)$
$\Rightarrow x-iy=\left(a^3-3a{b}^2\right)-i\left(b^3-3a^{2}b\right)$
$[\because z=x-iy]$
Here, $x=a^3-3a{b}^2$ and $y=b^3-3a^{2}b$
Now, $\frac{\left(\frac{x}{a}+\frac{y}{b}\right)}{a^2+b^2}=\frac{\left(\frac{a^3-3a{b}^2}{a}+\frac{b^3-3a^{2}b}{b}\right)}{a^2+b^2}$
$=\frac{a^2-3b^2+b^2-3a^2}{a^2+b^2}$
$=\frac{-2a^2-2b^2}{a^2+b^2}=\frac{-2\left(a^2+b^2\right)}{a^2+b^2}=-2$