# If a series converges absolutely over the real numbers, then does it absolutely converge over the complex numbers too?

It was stated at the beginning of my lecture notes that the Taylor expansion of the sine, cosine, and exponential function converge, i. e. we have

$$\sin{x} = \sum_{\nu = 0}^\infty (-1)^\nu \frac{x^{2\nu+1}}{(2\nu+1)!}, \quad \cos{x} = \sum_{\nu = 0}^\infty (-1)^\nu \frac{x^{2\nu}}{(2\nu)!}, \quad \exp{x} = \sum_{\nu = 0}^\infty \frac{x^\nu}{\nu!}$$

for all $$x \in \mathbb{R}$$.

Then my notes follow up by saying that since these series converge absolutely, then they also converge absolutely for an arbitrary $$z \in \mathbb{C}$$. Thus the values of

$$\sin{z} := \sum_{\nu = 0}^\infty (-1)^\nu \frac{z^{2\nu+1}}{(2\nu+1)!}, \quad \cos{z} = \sum_{\nu = 0}^\infty (-1)^\nu \frac{z^{2\nu}}{(2\nu)!}, \quad \exp{z} = \sum_{\nu = 0}^\infty \frac{z^\nu}{\nu!},$$

are well-defined.

My question: Why is the part that I highlighted in italics true? Does the argument only work for $$\sin, \cos, \exp$$ or is there a more general argument?

• Take for example $\exp{z} = \sum\limits_{\nu = 0}^\infty \frac{z^\nu}{\nu!}$. This will converge absolutely if $\sum\limits_{\nu = 0}^\infty \left|\frac{z^\nu}{\nu!}\right| =\sum\limits_{\nu = 0}^\infty \frac{|z|^\nu}{\nu!}$ converges, and it does since $\sum\limits_{\nu = 0}^\infty \frac{x^\nu}{\nu!}$ converges absolutely for $x=|z|$. Sep 24 at 23:53
• It works for all power series, i.e. series of the type $\sum a_nz^{n}$. Sep 24 at 23:53

It holds true for all power series.

Consider the series $$A(z) = \sum a_n z^n$$, and the complex number $$z = r e^{i\theta}$$. Then if we look at the absolute convergence we have

$$\begin{eqnarray} \sum |a_n z^n| & = & \sum |a_n (r e^{i \theta})^n| \\ & = & \sum |a_n| |r^n| |e^{i n \theta}| \\ & = & \sum |a_n| r^n \end{eqnarray}$$

since $$r \geq 0$$ and $$|e^{i n \theta}| = 1$$. So if the series converges absolutely on the reals, then the last term in the above equation converges absolutely, meaning the series in general converges.

• Not all series, but all power series. Sep 25 at 4:50
• Thanks, have fixed. Sep 25 at 7:08

Let $$P(x) = \sum_{n=0}^\infty a_nx^n$$ be a real power series (this means that all $$a_n \in \mathbb R$$ and $$x$$ is a real variable) and $$P(z) = \sum_{n=0}^\infty a_nz^n$$ the "extended" complex power series with a complex variable $$z$$.

Absolute converge of $$P(x)$$ for all $$x \in \mathbb R$$ means that $$\sum_{n=0}^\infty \lvert a_n x^n \rvert = \sum_{n=0}^\infty \lvert a_n \rvert \cdot \lvert x \rvert^n$$ is convergent for all $$x \in \mathbb R$$, and this is equivalent to the convergence of $$\sum_{n=0}^\infty \lvert a_n \rvert \cdot r^n$$ for all $$r \ge 0$$.

Similarly absolute converge of $$P(z)$$ for all $$z \in \mathbb C$$ means that $$\sum_{n=0}^\infty \lvert a_n z^n \rvert = \sum_{n=0}^\infty \lvert a_n \rvert \cdot \lvert z \rvert^n$$ is convergent for all $$z\in \mathbb C$$, and again this is equivalent to the convergence of $$\sum_{n=0}^\infty \lvert a_n \rvert \cdot r^n$$ for all $$r \ge 0$$.

This shows that absolute converge of $$P(x)$$ for all $$x \in \mathbb R$$ is equivalent to absolute converge of $$P(z)$$ for all $$z \in \mathbb C$$.

The general context is this: Each real or complex power series $$\Pi(w) = \sum_{n=0}^\infty \alpha_n w^n$$ (here the $$\alpha_n$$ are real coefficients and $$w$$ is a real variable, or the $$\alpha_n$$ are complex coefficients and $$w$$ is a complex variable) has radius of convergence $$R \ge 0$$ which is given by $$R = \frac{1}{\limsup_{n \to \infty}\sqrt[n]{\lvert \alpha_n \rvert}} . \tag{1}$$ $$\Pi(w)$$ is absolutely convergent for $$\lvert w \rvert < R$$ and divergent for $$\lvert w \rvert > R$$.

You see that formula $$(1)$$ is independent on the ground field ($$\mathbb R$$ or $$\mathbb C$$), it only depends on the abolutes values $$\lvert \alpha_n \rvert$$ of the coefficients.

In case of the power series for $$\sin, \cos$$ and $$\exp$$ we have $$R = \infty$$.