I know by definition that the radius of convergence is $R:=\sup\{|z|\in\mathbb{R}\colon\sum_{n=0}^{\infty} a_n z^n \text{ converges}\}$
I don't understand why
$R:=\sup\{z\in\mathbb{C}\colon\sum_{n=0}^{\infty} a_n z^n \text{ converges}\}$ it is not correct since $z\in\mathbb{C}$
On
The supremum of a set $A \subset B$ is defined in terms the ordering of the elements of $B$: for two elements $x\in B$ and $y\in B$ we need to be able to say whether $x\leq y$.
For real numbers, $\leq$ is defined, and if $A\subset \mathbb R$ then $A$ has a supremum in $\mathbb R$.
For the complex numbers, no such ordering exists and $x \leq y$ has no meaning, so the supremum of a set of complex numbers has no meaning either.
The clue, though, is in the word radius. This is a distance. $|z|$ is the distance of $z$ from the origin. In your example, $R$ is defined in terms of that—and convergence will in fact happen inside an actual circle of radiius $R$ on the complex plane.
Because the set$$\left\{z\in\mathbb{C}\,\middle|\,\sum_{n=0}^\infty a_nz^n\text{ converges}\right\}\tag1$$either is $\{0\}$ or it contains complex non-real numbers. In the later case, since there is no order relation in $\mathbb C$, it makes no sense to talk about the supremum of $(1)$.