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?

  • 1
    $\begingroup$ 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 $x=|z|$. $\endgroup$
    – Henry
    Sep 24 at 23:53
  • $\begingroup$ It works for all power series, i.e. series of the type $\sum a_nz^{n}$. $\endgroup$ Sep 24 at 23:53

2 Answers 2


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.

  • $\begingroup$ Not all series, but all power series. $\endgroup$ Sep 25 at 4:50
  • $\begingroup$ Thanks, have fixed. $\endgroup$
    – ConMan
    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 $R = \infty$.


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