Question

If x + iy = (a + ib)³, show that $\frac{x}{a}+\frac{y}{b}=4(a^{2}-b^{2})$.

Answer 1

Answer:

Given:

x + iy = (a + ib)³.

RHS of the equation:

$\rm (a+ib)^3=(a+ib)^2(a+ib)\\=(a^2+(ib)^2+2a(ib))(a+ib)\\=(a^2+(-b^2)+2iab)(a+ib)\ \ \ \ \ \because (i^2=-1)\\=(a^2-b^2+2iab)(a+ib)\\=(a^3-b^2a+2ia^2b)+(a^2ib-ib^3+2i^2ab^2)\\=a^3-ab^2+i2a^2b+ia^2b-ib^3-2ab^2\\=(a^3-ab^2-2ab^2)+i(2a^2b+a^2b-b^3)\\=a(a^2-b^2-2b^2)+ib(2a^2+a^2-b^2)\\=a(a^2-3b^2)+ib(3a^2-b^2)$

On comparing LHS with RHS,

$\rm x+iy=a(a^2-3b^2)+ib(3a^2-b^2)\\x=a(a^2-3b^2)\\y=b(3a^2-b^2)\\\\\dfrac xa=a^2-3b^2\\\dfrac yb = 3a^2-b^2\\\dfrac xa +\dfrac yb=(a^2-3b^2) +( 3a^2-b^2) = 4a^2-4b^2=4(a^2-b^2).$

Hence, showed that LHS = RHS.