Question

If (x + iy)1/3 = a + ib, then $\frac{x}{a}+\frac{y}{b}=$
(a) 0
(b) 1
(c) −1
(d) none of these

Answer 1

(d) none of these

$$\begin{gathered} {\left(x+iy\right)}^{\frac{1}{3}}=a+ib \\ \mathrm{Cubing}\mathrm{on}\mathrm{both}\mathrm{the}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ x+iy={\left(a+ib\right)}^3 \\ \Rightarrow x+iy=a^3+{\left(ib\right)}^3+3a^{2}bi+3a{\left(ib\right)}^2 \\ \Rightarrow x+iy=a^3+i^3{b}^3+3a^{2}ib+3i^{2}a{b}^2 \\ \Rightarrow x+iy=a^3-i{b}^3+3a^{2}ib-3a{b}^2(\because i^2=-1,i^3=-i) \\ \Rightarrow x+iy=a^3-3a{b}^2+i\left(-b^3+3a^{2}b\right) \\ \therefore x=a^3-3a{b}^2\mathrm{and}y=3a^{2}b-b^3 \\ \mathrm{or},\frac{x}{a}=a^2-3b^2\mathrm{and}\frac{y}{b}=3a^2-b^2 \\ \Rightarrow \frac{x}{a}+\frac{y}{b}=a^2-3b^2+3a^2-b^2 \\ \Rightarrow \frac{x}{a}+\frac{y}{b}=4a^2-4b^2 \end{gathered}$$