Why radius of convergence of a power series can only be real

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 $$z\in\mathbb{C}$$

• Because the modulus of a complex number is a non-negative real number. – Bernard Jan 6 at 13:25
• The supremum of an arbitrary set of complex numbers is not defined. To talk about suprema, you need an ordering. – Andrés E. Caicedo Jan 6 at 13:34
• 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. – jgon Jan 10 at 1:34

2 Answers

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 $$(1)$$.

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.