Question

${\left(x+iy\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=a+ib$, then $\left(\frac{x}{a}\right)+\left(\frac{y}{b}\right)$ is equal to

• A. $4\left(a^2+b^2\right)$
• B. $4\left(a^2-b^2\right)$
• C. $4\left(b^2-a^2\right)$
• D. None of these.

Answer 1

The correct option is B

$4\left(a^2-b^2\right)$

Explanation for the correct option

Step 1: Note the given data

The equation is ${\left(x+iy\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=a+ib$

Taking the cube on both sides,

$\Rightarrow$${\left[{\left(x+iy\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right]}^3={\left(a+ib\right)}^3$

$\Rightarrow$ $\left(x+iy\right)=a^3+3a^2·ib+3a{\left(ib\right)}^2+{\left(ib\right)}^3\left[\because \mathrm{using}\mathrm{the}\mathrm{identity}{\left(x+y\right)}^3=x^3+3x^{2}y+3x{y}^2+y^3\right]$

$\Rightarrow$ $\left(x+iy\right)=\left(a^3-3a{b}^2\right)+\left(3a^{2}b-b^3\right)i\left[\because \text{grouping the real and imaginary part}\right]$

Step 2: Compare the equation to write the values of $x$and $y$

$\Rightarrow$ $x=a^3-3a{b}^{2}andy=3a^{2}b-b^3$

$\Rightarrow$ $x=a\left(a^2-3b^2\right)andy=b\left(3a^2-b^2\right)$

$\Rightarrow$ $\frac{x}{a}=\left(a^2-3b^2\right)and\frac{y}{b}=\left(3a^2-b^2\right)$

Step 3: Find the value of $\left(\frac{x}{a}\right)+\left(\frac{y}{b}\right)$

$\left(\frac{x}{a}\right)+\left(\frac{y}{b}\right)=\left(a^2-3b^2\right)+\left(3a^2-b^2\right)$

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

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

Hence, the correct option is B.