Question

lf (x-iy) 1/3 =a+ib then x/a ​ + y/b ​ =​

Answer 1

$\large\underline{\sf{Solution-}}$

Given complex number is

$\rm :\longmapsto\: {\bigg(x - iy\bigg)}^{\dfrac{1}{3} } = a + ib$

Cubing both sides, we get

$\rm :\longmapsto\:x - iy = {(a + ib)}^{3}$

We know,

$\boxed{ \tt{ \: {(x + y)}^{3} = {x}^{3} + {3x}^{2}y + {3xy}^{2} + {y}^{3} \: }}$

So, using this identity, we get

$\rm :\longmapsto\:x - iy = {a}^{3} + 3 {a}^{2} (ib) + 3a {(ib)}^{2} + {(ib)}^{3}$

$\rm :\longmapsto\:x - iy = {a}^{3} + 3 {a}^{2}bi + 3a {b}^{2} {i}^{2} + {i}^{3} {b}^{3}$

We know,

$\boxed{ \tt{ \: {i}^{2} \: = \: - \: 1 \: }} \: \: and \: \: \boxed{ \tt{ \: {i}^{3} \: = \: - \: i \: }}$

So, using these, we get

$\rm :\longmapsto\:x - iy = {a}^{3} + 3 {a}^{2}bi - 3a {b}^{2} - i{b}^{3}$

$\rm :\longmapsto\:x - iy = {a}^{3} - 3 {ab}^{2} + 3 {a}^{2}bi - i{b}^{3}$

$\rm :\longmapsto\:x - iy = a( {a}^{2} - {3b}^{2}) +ib( 3 {a}^{2}- {b}^{2})$

So, On comparing real and Imaginary parts, we get

$\rm :\longmapsto\:x = a( {a}^{2} - {3b}^{2}) - - - - (1)$

and

$\rm :\longmapsto\:y = - b( {3a}^{2} - {b}^{2}) - - - - (2)$

Now, Consider

$\rm :\longmapsto\:\dfrac{x}{a} + \dfrac{y}{b}$

On substituting the values of x and y from equation (1) and equation (2), we get

$\rm \: = \: \dfrac{a( {a}^{2} - {3b}^{2})}{a} - \dfrac{b( {3a}^{2} - {b}^{2} )}{b}$

$\rm \: = \: {a}^{2} - {3b}^{2} - ( {3a}^{2} - {b}^{2})$

$\rm \: = \: {a}^{2} - {3b}^{2} - {3a}^{2} + {b}^{2}$

$\rm \: = \: - {2b}^{2} - {2a}^{2}$

$\rm \: = \: - 2( {a}^{2} + {b}^{2})$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{x}{a} + \dfrac{y}{b} = 2( {a}^{2}+{b}^{2})}}$