If z=x-iy and z^{1/3}=a+ib,then frac{(x/a+y/b | vedclass

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

Question

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

Vedclass · Accepted Answer

-$2$

Vedclass · Answer

(A) Given,$z=x-iy$ and $z^{1/3}=a+ib$. Taking the cube on both sides,we get: $(z^{1/3})^3 = (a+ib)^3$ $z = a^3 + (ib)^3 + 3a^2(ib) + 3a(ib)^2$ $z = a^3 - ib^3 + 3a^2bi - 3ab^2$ $z = (a^3 - 3ab^2) - i(b^3 - 3a^2b)$ Since $z = x - iy$,we compare the real and imaginary parts: $x = a^3 - 3ab^2$ and $y = b^3 - 3a^2b$. Now,substitute these into the expression: $\frac{(x/a + y/b)}{a^2+b^2} = \frac{((a^3 - 3ab^2)/a + (b^3 - 3a^2b)/b)}{a^2+b^2}$ $= \frac{(a^2 - 3b^2 + b^2 - 3a^2)}{a^2+b^2}$ $= \frac{-2a^2 - 2b^2}{a^2+b^2} = \frac{-2(a^2+b^2)}{a^2+b^2} = -2$.

If $z=x-iy$ and $z^{1/3}=a+ib$,then $\frac{(x/a+y/b)}{a^2+b^2}=$

Similar Questions

If $z_1, z_2, z_3, z_4$ are the roots of the equation $z^4 + z^3 + z^2 + z + 1 = 0$,then $\prod_{i=1}^{4} (z_i + 2)$ is equal to:

Let $z$ and $w$ be two distinct non-zero complex numbers. If $|z|^2 w - |w|^2 z = z - w$,then:

Find the complex number $z$ satisfying the equations $\left| \frac{z - 12}{z - 8i} \right| = \frac{5}{3}$ and $\left| \frac{z - 4}{z - 8} \right| = 1$.

If $z$ is a complex number such that $\left|z-\frac{4}{z}\right|=2$,then the greatest value of $|z|$ is

The number of all possible solutions of the equation $z^3+\overline{z}=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module