Question

lf ${\left(x+iy\right)}^{\frac{1}{3}}=a+ib$ then $\frac{x}{a}+\frac{y}{b}=$?

• A. $a^2-b^2$
• B. $2(a^2-b^2)$
• C. $3(a^2-b^2)$
• D. $4(a^2-b^2)$

Answer 1

The correct option is D

$4(a^2-b^2)$

Explanation for the correct option:

Step 1: Simplify the term ${\left(x+iy\right)}^{\frac{1}{3}}=a+ib$:

Given: ${\left(x+iy\right)}^{\frac{1}{3}}=a+ib$

$$\begin{gathered} {\left(x+iy\right)}^{\frac{1}{3}}=a+ib \\ \left(x+iy\right)={\left(a+ib\right)}^3 \\ \left(x+iy\right)=a^3+{\left(ib\right)}^3+3a^{2}ib+3a{b}^2{i}^2 \\ (x+iy)=a^3-i{b}^3+i3a^{2}b-3a{b}^2 \\ (x+iy)=a^3-3a{b}^2-i{b}^3+i3a^{2}b \\ x+iy=\left(a^3-3a{b}^2\right)+i\left(-b^3+3a^{2}b\right) \end{gathered}$$

comparing like terms

$x=a^3-3a{b}^2,y=-b^3+3a^{2}b$

Step 2: Put the value of $x$and $y$ in the equation $\frac{x}{a}+\frac{y}{b}$:

$$\begin{gathered} \frac{x}{a}+\frac{y}{b}=\frac{a^3-3a{b}^2}{a}+\frac{-b^3+3a^{2}b}{b} \\ \frac{x}{a}+\frac{y}{b}=a^2-3b^2-b^2+3a^2 \\ \frac{x}{a}+\frac{y}{b}=4a^2-4b^2 \\ \frac{x}{a}+\frac{y}{b}=4\left(a^2-b^2\right) \end{gathered}$$

Hence, the correct option is (D)