Question

Solve it


If ( x + iy)⅓ = a + ib ( x , y , a , b £ R )

Show that x/a + y/b = 4 (a² - b²)​

Answer 1

Heya!!

_______________________________________________________________

$<b> Complex Number Systems$

Given that ,

$= > (x +i y) {}^{ \frac{1}{3} } = a + ib \\ \\ = > (x + iy) = (a + ib) {}^{3}$

=> x + iy = a³ + 3a²ib + 3a i²b² + i³b³

=> x + iy = ( a³ - 3ab²) + i (3a²b - b³)

$= > \frac{x}{a} = a {}^{2} - 3b {}^{2} \: and \: \frac{y}{b} = 3a {}^{2} - b {}^{2}$

$<u> Adding the two values </u>$

$= > \frac{x}{a} + \frac{y}{b} = a {}^{2} - 3b {}^{2} + 3a {}^{2} - b {}^{2}$

$= \frac{x}{a} + \frac{y}{b} = 4(a {}^{2} - b {}^{2} )$

L.H.S = R.H.S

$<b> Hence Proved!! </b>$

____________________________________________________________

Answer 2

Hello!

( x + iy)^1/3 = (a + ib)

So cubing both the sides

( x + iy) = (a+ib)³

( x + iy) = a³ + i³b³ + 3a²ib + 3ai²b²

Then You can separate the real and imaginary parts

x = a³ - 3ab²

y = b³ + 3a²b

Put the values x/a + y/b

= 4a² - 4b²

= 4 (a² - b²)

Hope it the the correct answer friend.☺

Hope it is helpful