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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 $z\in\mathbb{C}$

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    $\begingroup$ Because the modulus of a complex number is a non-negative real number. $\endgroup$ – Bernard Jan 6 at 13:25
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    $\begingroup$ The supremum of an arbitrary set of complex numbers is not defined. To talk about suprema, you need an ordering. $\endgroup$ – Andrés E. Caicedo Jan 6 at 13:34
  • $\begingroup$ I have to say I think the people voting to close this for lack of context are being deceived by its brevity. I think the first sentence gives sufficient context to understand the motivation for this question, and the confusion on part of the asker. $\endgroup$ – jgon Jan 10 at 1:34
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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 $(1)$.

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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.

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