If (x+iy)^{1/3} = a+ib where x, y, a, b in R | 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 $x+iy = (a+ib)^3$.Using the identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,we have $x+iy = a^

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $(x+iy)^{1/3} = a+ib$.Cubing both sides,we get $x+iy = (a+ib)^3$.Using the identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,we have $x+iy = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3$.$x+iy = a^3 + 3a^2bi - 3ab^2 - ib^3$.Grouping real and imaginary parts: $x+iy = (a^3 - 3ab^2) + i(3a^2b - b^3)$.Equating real and imaginary parts: $x = a^3 - 3ab^2$ and $y = 3a^2b - b^3$.Dividing by $a$ and $b$ respectively: $\frac{x}{a} = a^2 - 3b^2$ and $\frac{y}{b} = 3a^2 - b^2$.Therefore,$\frac{x}{a} - \frac{y}{b} = (a^2 - 3b^2) - (3a^2 - b^2) = a^2 - 3b^2 - 3a^2 + b^2 = -2a^2 - 2b^2 = -2(a^2+b^2)$.

If $(x+iy)^{1/3} = a+ib$ where $x, y, a, b \in R$ and $i = \sqrt{-1}$,then $\frac{x}{a} - \frac{y}{b} = $

Similar Questions

Let $z$ be a complex number satisfying $|z - 5i| \le 1$ such that $\text{amp } z$ is minimum. Then $z$ is equal to

Consider the regions for complex number $z$ defined by $A: \frac{1}{\log_2 |z|} - \frac{1}{\log_2 |z| - 1} - 1 < 0$ and $B: \operatorname{Im}(z) = 0$. The range of values of $\operatorname{Re}(z)$ lying in the region $A \cap B$ is

If $x = -5 + 2 \sqrt{-4}$,then the value of $x^4 + 9x^3 + 35x^2 - x + 4$ is

If $\cos A+\cos B+\cos C=0$ and $\sin A+\sin B+\sin C=0$,then $\cos (A-B)=$

Let $z_1, z_2 \in \mathbb{C}$ be the distinct solutions of the equation $z^2 + 4z - (1 + 12i) = 0$. Then $|z_1|^2 + |z_2|^2$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module