The number of all possible solutions of the e | vedclass

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

Question

(B) Given the equation $z^3+\bar{z}=0$. Taking the modulus on both sides,we get $|z^3| = |-\bar{z}|$,which implies $|z|^3 = |z|$. This gives $|z|(|z|^

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Given the equation $z^3+\bar{z}=0$. Taking the modulus on both sides,we get $|z^3| = |-\bar{z}|$,which implies $|z|^3 = |z|$. This gives $|z|(|z|^2 - 1) = 0$. Case $1$: $|z| = 0$,which implies $z = 0$. This is $1$ solution. Case $2$: $|z|^2 = 1$,which implies $z\bar{z} = 1$,so $\bar{z} = \frac{1}{z}$. Substituting this into the original equation: $z^3 + \frac{1}{z} = 0$,which gives $z^4 + 1 = 0$. The equation $z^4 = -1$ has $4$ distinct roots. Thus,the total number of solutions is $1 + 4 = 5$.

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

Similar Questions

The minimum degree of a polynomial equation with rational coefficients having $\sqrt{3}+\sqrt{27}$ and $\sqrt{2}+5i$ as two of its roots is

If $x=p+q$,$y=p \omega+q \omega^2$ and $z=p \omega^2+q \omega$,where $\omega$ is a complex cube root of unity,then $xyz$ is equal to

If $z = x + iy$ $(x, y \in R, x \neq -1/2)$,the number of values of $z$ satisfying $|z|^n = z^2|z|^{n-2} + z|z|^{n-2} + 1$ $(n \in N, n > 1)$ is

$\frac{(1+i)^{2016}}{(1-i)^{2014}}$ is equal to

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module