If z = x - iy and z^{1/3} = p + iq,then left( | vedclass

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

Question

(A) Given $z = x - iy$ and $z^{1/3} = p + iq$.Taking the cube on both sides,we get $z = (p + iq)^3$.$z = p^3 + 3p^2(iq) + 3p(iq)^2 + (iq)^3$.$z = p^3

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(A) Given $z = x - iy$ and $z^{1/3} = p + iq$.Taking the cube on both sides,we get $z = (p + iq)^3$.$z = p^3 + 3p^2(iq) + 3p(iq)^2 + (iq)^3$.$z = p^3 + 3ip^2q - 3pq^2 - iq^3$.$z = (p^3 - 3pq^2) + i(3p^2q - q^3)$.Comparing the real and imaginary parts with $z = x - iy$,we have:$x = p^3 - 3pq^2 = p(p^2 - 3q^2)$ and $-y = 3p^2q - q^3$,which implies $y = q^2 - 3p^2q = q(q^2 - 3p^2)$.Now,$\frac{x}{p} = p^2 - 3q^2$ and $\frac{y}{q} = q^2 - 3p^2$.Adding these two expressions:$\frac{x}{p} + \frac{y}{q} = (p^2 - 3q^2) + (q^2 - 3p^2) = -2p^2 - 2q^2 = -2(p^2 + q^2)$.Therefore,$\frac{\frac{x}{p} + \frac{y}{q}}{p^2 + q^2} = \frac{-2(p^2 + q^2)}{p^2 + q^2} = -2$.

If $z = x - iy$ and $z^{1/3} = p + iq$,then $\left( \frac{x}{p} + \frac{y}{q} \right) / (p^2 + q^2)$ is equal to

Similar Questions

For a complex number $Z = a + ib$,let $\hat{Z} = b + ia$. If $Z_1$ and $Z_2$ are such complex numbers,then $\widehat{Z_1 Z_2} = $

Let $S = \{z \in \mathbb{C} : \bar{z} = i(z^2 + \operatorname{Re}(\bar{z}))\}$. Then $\sum_{z \in S} |z|^2$ is equal to

If $2i$ is a root of $f(z) = z^4 + z^3 + 2z^2 + 4z - 8 = 0$,then which among the following cannot be a root of $f(z) = 0$?

The value of $\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^3$ is

$z_1$ and $z_2$ are two complex numbers such that $|z_1 + z_2| = 1$ and $|z_1^2 + z_2^2| = 25$. Then the minimum value of $|z_1^3 + z_2^3|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module