Question

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

• A. 0
• B. 1
• C. -1
• D. none of these

Answer 1

The correct option is D

none of these

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

Cubing on both the sides, we get :

$x+iy=(a+ib{)}^3\Rightarrow x+iy=a^3+(ib{)}^3+3a^{2}bi+3a(ib{)}^2\Rightarrow x+iy=a^3+i^3{b}^3+3a^{2}ib+3a{i}^2{b}^2\Rightarrow x+iy=a^3-i{b}^3+3a^{2}ib-3a{b}^2(\because i^2=-1,i^3=-i)\Rightarrow x+iy=a^3-3a{b}^2+i(-b^3+3a^{2}b)\therefore x=a^3-3a{b}^2\text{and}y=3a^{2}b-b^{3}or,\frac{x}{a}=a^2-3b^2\text{and}\frac{y}{b}=3a^2-b^2\Rightarrow \frac{x}{a}+\frac{y}{b}=a^2-3b^2+3a^2-b^2\Rightarrow \frac{x}{a}+\frac{y}{b}=4a^2-4b^2$