Question

if z=a+iband z³=x+iy then find

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Answer 1

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

Given that,

$\rm :\longmapsto\:z = a + ib$

and

$\rm :\longmapsto\: {z}^{3} = x + iy$

On substituting the value of z, we get

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

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

We know,

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

So, using this identity, we get

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

We know,

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

Using this, we get

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

can be re-arranged as

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

So, on comparing, we get

$\red{\rm :\longmapsto\:x = {a}^{3} - {3ab}^{2}}$

and

$\red{\rm :\longmapsto\:y = {3ba}^{2} - {b}^{3} }$

Now, Consider

$\rm :\longmapsto\:\dfrac{1}{ {a}^{2} - {b}^{2} }\bigg[\dfrac{x}{a} + \dfrac{y}{b} \bigg]$

$\rm \:  =  \: \dfrac{1}{ {a}^{2} - {b}^{2} }\bigg[\dfrac{ {a}^{3} - 3 {ab}^{2} }{a} + \dfrac{ {3ba}^{2} - {b}^{3} }{b} \bigg]$

$\rm \:  =  \: \dfrac{1}{ {a}^{2} - {b}^{2} }\bigg[\dfrac{a({a}^{2} - 3 {b}^{2})}{a} + \dfrac{b({3a}^{2} - {b}^{2})}{b} \bigg]$

$\rm \:  =  \: \dfrac{1}{ {a}^{2} - {b}^{2} }\bigg[ {a}^{2} - {3b}^{2} + {3a}^{2} - {b}^{2} \bigg]$

$\rm \:  =  \: \dfrac{1}{ {a}^{2} - {b}^{2} }\bigg[ 4{a}^{2} - {4b}^{2} \bigg]$

$\rm \:  =  \: \dfrac{4}{ {a}^{2} - {b}^{2} }\bigg[ {a}^{2} - {b}^{2} \bigg]$

$\rm \:  =  \: 4$

Hence,

$\rm \implies\:\boxed{ \tt{ \: \dfrac{1}{ {a}^{2} - {b}^{2} }\bigg[\dfrac{x}{a} + \dfrac{y}{b} \bigg] = 4 \: }}$

Answer 2

Answer:

Step-by-step explanation:

We have,

$\tt{z=a+ib\,\,\,\,\,\,and\,\,\,\,\,\,\,z^3=x+iy}$

$\sf{\implies\,z^3=(a+ib)^3}$

$\sf{\implies\,x+iy=(a)^3+(ib)^3+3(a)^2(ib)+3(a)(ib)^2}$

$\sf{\implies\,x+iy=a^3-ib^3+3a^2b\,i-3ab^2}$

$\sf{\implies\,x+iy=a^3-3ab^2+(3a^2b-b^3)i}$

On comparing

$\sf{\implies\,x=a^3-3ab^2\,\,\,\,\,\,\,\&\,\,\,\,\,\,\,y=3a^2b-b^3}$

$\sf{\implies\,x=a(a^2-3b^2)\,\,\,\,\,\,\,\&\,\,\,\,\,\,\,y=b(3a^2-b^2)}$

$\sf{\implies\,\dfrac{x}{a}=a^2-3b^2\,\,\,\,\,\,\,\&\,\,\,\,\,\,\,\dfrac{y}{b}=3a^2-b^2}$

Now,

$\sf{\dfrac{x}{a}+\dfrac{y}{b}=a^2-3b^2+3a^2-b^2}$

$\sf{\implies\dfrac{x}{a}+\dfrac{y}{b}=4a^2-4b^2}$

$\sf{\implies\dfrac{1}{a^2-b^2}\cdot\left(\dfrac{x}{a}+\dfrac{y}{b}\right)=4}$