If (x + iy)3 = u + iv, then show that $$\dfrac{u}{x}+\dfrac{v}{y}=4(x^2-y^2)$$

Asked by Abhisek | 1 year ago |  134

##### Solution :-

(x + iy)3 = u + iv

x3 + (iy)3 + 3× x ×iy(x + iy) = u + iv

x3 + i3y3 + 3x2yi + 3xy2i2 = u + iv

x3 - iy3 + 3x2yi - 3xy2 = u + iv

(x3 - 3xy2) + i(3x2y - y3) = u + iv

On equating real and imaginary

u = x3 - 3xy2 ,v = 3x2y - y3

$$\dfrac{u}{x}+\dfrac{v}{y}=\dfrac{x^3-3xy^2}{x}+\dfrac{3x^2y-y^3}{y}$$

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

x2 - 3y2 + 3x2 - y

4x2 - 4y2

4(x- y2)

Hence, proved

Answered by Pragya Singh | 1 year ago

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