Why does $z^3=0$ have 3 complex roots and not simply just $0$?

By the fundamental theorem of algebra, it should be that $z^3=0$ has 3 solutions. However, when I attempt to calculate the roots,

$$z^3=0\iff (re^{i\theta})^3=0e^{i\theta_1}$$ so $3i\theta =i\theta_1$ which implies that $\theta=\frac{\theta_1+2\pi ik}{3}$ where $k\in\{0,1,2\}$ and $r=0$. However, since the radius is zero, wouldn't all three roots simply be zero meaning it only has a single root which is $0+0bi$?

The fundamental theorem of algebra states that every polynomial of degree $n$ can be written as $c(z-z_1)(z-z_2)...(z-z_n)$ where $z_1, z_2,...,z_n$ are not necessary distinct. If the root $z_i$ appears exactly $k$ times in this decomposition then we say this root has multiplicity $k$. So if we count such a root $k$ times then the polynomial has exactly $n$ solutions.

For example, the polynomial $p(z)=z^3$ has just one root, and the multiplicity of this root is $3$.