Question

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

Answer 1

Please see the attachment

Answer 2

Answer:

u/x + v/y = 4(x² - y²)

Step-by-step explanation:

(x + iy)³

Using (a + b)³ = a³ + b³ + 3a²b + 3ab²

a = x  b = iy

= x³ + (iy)³ + 3x²(iy) + 3x(iy)²

i³ = -i  & i² = -1

= x³ - iy³ + i3x²y - 3xy²

= x³ - 3xy² + i3x²y - iy³

= x(x² - 3y²)  + iy(3x² - y²)

Comparing with

u + iv

u =x(x² - 3y²) => u/x = x² - 3y²

v = y(3x² - y²) => v/y = 3x² - y²

adding both

u/x + v/y = x² - 3y² + 3x² - y²

=> u/x + v/y =  4x² - 4y²

=> u/x + v/y = 4(x² - y²)