If (x-iy)^{1/3} = a-ib,then the value of frac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $(x-iy)^{1/3} = a-ib$. Cubing both sides,we get: $x-iy = (a-ib)^3$ $x-iy = a^3 - (ib)^3 - 3a^2(ib) + 3a(ib)^2$ $x-iy = a^3 + ib^3 - 3a^2bi -

Vedclass · Accepted Answer

$4(a^2-b^2)$

Vedclass · Answer

(A) Given $(x-iy)^{1/3} = a-ib$. Cubing both sides,we get: $x-iy = (a-ib)^3$ $x-iy = a^3 - (ib)^3 - 3a^2(ib) + 3a(ib)^2$ $x-iy = a^3 + ib^3 - 3a^2bi - 3ab^2$ $x-iy = (a^3 - 3ab^2) - i(3a^2b - b^3)$ Comparing real and imaginary parts: $x = a^3 - 3ab^2 \implies \frac{x}{a} = a^2 - 3b^2$ $y = 3a^2b - b^3 \implies \frac{y}{b} = 3a^2 - b^2$ Adding these: $\frac{x}{a} + \frac{y}{b} = (a^2 - 3b^2) + (3a^2 - b^2)$ $= 4a^2 - 4b^2 = 4(a^2 - b^2)$

If $(x-iy)^{1/3} = a-ib$,then the value of $\frac{x}{a} + \frac{y}{b}$ is

Similar Questions

If $1+2i$ is a root of the equation $x^4-3x^3+8x^2-7x+5=0$,then the sum of the squares of the other roots is

If $(2+i)$ is a root of the equation $x^3-5x^2+9x-5=0$,then the other roots are

The complex numbers $\sin x + i\cos 2x$ and $\cos x - i\sin 2x$ are conjugate to each other for

If $x = 3 + i$,then $x^3 - 3x^2 - 8x + 15 = $

Let $\alpha$ and $\beta$ be the sum and the product of all the non-zero solutions of the equation $(\bar{z})^2+|z|=0$,where $z \in \mathbb{C}$. Then $4(\alpha^2+\beta^2)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module