Question

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

Answer 1

Solve for $\frac{x}{a}+\frac{y}{b}$

It is given that ${\left(x+iy\right)}^{\frac{1}{3}}=a+ib$

$$\begin{gathered} \Rightarrow \left(x+iy\right)={\left(a+ib\right)}^3 \\ \Rightarrow \left(x+iy\right)=a^3+{\left(ib\right)}^3+3abi\left(a+ib\right) \\ \Rightarrow \left(x+iy\right)=a^3-i{b}^3+3abi\left(a+ib\right) \\ \Rightarrow \left(x+iy\right)=a^3-i{b}^3+3a^{2}bi-3a{b}^2 \end{gathered}$$

Now compare L.H.S and R.H.S

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

Hence the value of $\frac{x}{a}+\frac{y}{b}$is $4\left(a^2-b^2\right)$