Question

If ${(x-iy)}^{\frac{1}{3}}=a-ib$ then show that $\frac{x}{a}+\frac{y}{b}=4(a^2-b^2)$

Answer 1

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

To prove:- $\frac{x}{a}+\frac{y}{b}=4(a^2-b^2)$

Proof:-

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

$x-iy={(a-ib)}^3$

$x-iy=a^3+i{b}^3-3a^{2}bi-3a{b}^2$

$x-iy=(a^3-3a{b}^2)-(3a^{2}b-b^3)i$

As we know that two complex numbers are equal if their corresponding parts are equal.

Therefore,

$x=a^3-3a{b}^2$

$\Rightarrow \frac{x}{a}=a^2-3b^2.....\left(1\right)$

$y=3a^{2}b-b^3$

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

Adding equation $\left(1\right)\&\left(2\right)$, we have

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

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

Hence proved.