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(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|^

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$5$

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(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

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