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

If $f(x)$ is a polynomial of degree $n$ with rational coefficients and $1+2i, 2-\sqrt{3}$ and $5$ are three roots of $f(x)=0$,then the least value of $n$ is

Let $z$ be a complex number such that $|z+2|=1$ and $\operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}$. Then the value of $|\operatorname{Re}(\overline{z+2})|$ is:

If $(x-iy)^{1/3} = 2-i\sqrt{3}$ and the point $z = (x, y)$ lies on the line $\frac{x}{2} + \frac{y}{\sqrt{3}} = k$,then $k =$

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

The number of solutions for $z^3+\bar{z}=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module