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kati45 [8]
3 years ago
5

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

Mathematics
2 answers:
V125BC [204]3 years ago
4 0

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

<u>Given that, </u>

\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

<u>We know, </u>

\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})

<u>Hence, Proved</u>

Elenna [48]3 years ago
3 0

Answer:

Step-by-step explanation:

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You can find the value of the hypotenuse if you apply the Pythagorean Theorem, which is show below:


 h²=a²+ b² ⇒ h=√(a² + b²)


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