Question

If (x+iy)³ = y+vi then show that (y/x + v/y) = 4(x²-y²)​

Answer 1

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

Given that,

$\rm \longmapsto\: {(x + iy)}^{3} = y + iv$

$\rm \longmapsto\: {x}^{3} + 3 {x}^{2}(iy) + 3x {(iy)}^{2} + {(iy)}^{3} = y + iv$

$\rm \longmapsto\: {x}^{3} + 3 {x}^{2}yi + 3x {i}^{2} {y}^{2} + {i}^{3} {y}^{3} = y + iv$

We know,

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

So, using this, we get

$\rm :\longmapsto\: {x}^{3} + 3 {x}^{2}yi - 3x{y}^{2} - i{y}^{3} = y + iv$

$\rm \longmapsto\: ({x}^{3} - 3x {y}^{2}) + i(3{x}^{2}y- {y}^{3})= y + iv$

On comparing real and Imaginary parts, we get

$\rm \longmapsto\:y = {x}^{3} - 3 {xy}^{2}$

and

$\rm \longmapsto\:v = {3x}^{2}y - {y}^{3}$

Now, Consider

$\rm \longmapsto\:\dfrac{y}{x} + \dfrac{v}{y}$

$\rm \: = \: \dfrac{ {x}^{3} - {3xy}^{2}}{x} + \dfrac{ {3yx}^{2} - {y}^{3} }{y}$

$\rm \: = \: \dfrac{ x({x}^{2} - {3y}^{2})}{x} + \dfrac{y({3x}^{2} - {y}^{2})}{y}$

$\rm \: = \: {x}^{2} - {3y}^{2} + {3x}^{2} - {y}^{2}$

$\rm \: = \: 4{x}^{2} - {4y}^{2}$

$\rm \: = \: 4({x}^{2} - {y}^{2})$

Hence, Proved

Answer 2

Answer:

Step-by-step explanation: