Some functions that can be differentiated indefinitely can be described ‘around each point’ as the sum of an power series. These are analytic functions, real or complex, the typical example being the exponential function, which can be extended to the whole complex plane.

1.Real and complex analytic functions

1.1.Developing the exponential function

We explained in ‘Deriving an inverse bijection & the example of the exponential function’ how to demonstrate that the exponential function $\exp:\mathbb R\to \mathbb R_+^*$, defined as the reciprocal bijection of the logarithm $\ln:\mathbb R_+^*\to \mathbb R$, is its own derivative. This means in particular that the function $\exp$ is indefinitely derivable. For such functions, known as ‘of class $\mathcal C^\infty$’, we can write the value of the function ‘around each point’ as what is known as a Taylor expansion, i.e. the sum of a polynomial of arbitrary degree, whose coefficients are given by the successive derivatives of the function, and a ‘remainder’.

In the case of the real exponential function and many other functions in real analysis, we have a stronger property: it is possible, ‘around any point’ $x_0$ of the domain of definition (here $\mathbb R$ for $\exp$), to give an expression of the function as a series whose coefficients are given by the successive derivatives of the function. For example, at the point $x_0=0$, we can describe the exponential function for any real number $x$ as $$\exp(x)=\sum_{n=0}^{+\infty} \dfrac{x^n}{n!}, $$ the property $\exp'(x)=\exp(x)$ allowing us to identify the coefficients of this series. The same thing can be done, for example, with the logarithm $\ln: \mathbb R_+^*\to \mathbb R$, this time with $x_0=1$, to obtain for any real number $x\in \mathbb R_+^*=]0,+\infty[$, $$\ln(x)=\sum_{n=1}^{+\infty} (-1)^{n+1}\frac{(x-1)^n}{n}. $$

1.2.Real and complex analytic functions

Functions for which it is possible to give such an expansion are said to be analytic. More formally, a function $f: I\to \mathbb C$, defined on an open interval $I$ of $\mathbb R$, is said to be analytic if for any point $x_0$ of $I$, there exists a sequence $(a_n)$ of complex numbers and an open interval $J\subseteq I$ such that $x_0\in J$ and for any $x\in J$, the complex series $\sum a_n (x-x_0)^n$ converges to $f(x)$. We say that $f$ is ‘expandable as a power series at (or around) $x_0$’. Thus, an analytic function is a function that can be uniformly described around each point as the sum of a particular series of polynomial functions.

If around each point $x_0\in I$, the coefficients of the sequence $(a_n)$ can be chosen to be real, then by definition the function $f$ takes its values in $\mathbb R$, and we say that we have a real analytic function $f:I\to \mathbb R$. On the other hand, in general for any $x_0\in I$ we can always choose an interval $J$ of the form $]x_0-\alpha,x_0+\alpha[=\{x\in \mathbb R : x_0-\alpha < x_0\}$. A detailed study would then show that for any complex number $z$ such that $|z-x_0|<\alpha$, the series $\sum a_n (z-x_0)^n$ converges and we can then extend the function $f$ at $z$ by putting $f(z)=\sum_{n=0}^{+\infty} a_n(z-x_0)^n$.

In fact, the notion of an analytic function is also valid for functions $f: U\to \mathbb C$, where $U$ is this time an open subset of $\mathbb C$, i.e. such that for any $z\in U$, there exists a real $\alpha >0$ such that the open ball $B_\alpha(z)=\{w\in \mathbb C : |z-w|<\alpha\}$, which contains $z$, is included in $U$. Such a function is said to be complex analytic, if for any $z\in U$, there exists an open ball $B_\alpha(z)\subseteq U$ and a sequence $(a_n)$ of complex numbers, such that for any $w\in B_\alpha(z)$, we have $f(w)=\sum_{n=0}^{+\infty} a_n (w-z)^n$.

1.3. Derivating an analytic function by its integer series

A real analytic function $f:I\to \mathbb R$ is always indefinitely derivable. Furthermore, at any point $x_0\in I$, if the function $f$ is described on an open interval $J\subseteq I$ by $f(x\in J)=\sum_{n=0}^{+\infty} a_n (x-x_0)^n$, its derivative is also described by a series, obtained from it in the same way as we derive a polynomial ‘term by term’. In other words, for all $x\in J$ we have $f'(x)=\sum_{n=1}^{+\infty} na_n (x-x_0)^{n-1}=\sum_{n=0}^{+\infty} (n+1)a_{n+1} (x-x_0)^n$. So the derivative of a real analytic function is still a real analytic function. The case of an analytic function $f:I\to \mathbb C$ is in fact similar: the derivative of such a function is well defined as that of a function with values in $\mathbb R^2$, and the same result holds.

In the complex case, for a function $f:U\to \mathbb C$ defined on an open subset $U$ of $\mathbb C$, we can define the complex derivability of $f$ at a point $z\in U$, in a similar way to the real definition. We say that $f$ is differentiable or derivable at $z$ if the ratio $\dfrac{f(w)-f(z)}{w-z}$ has a limit on $U$, when $w$ tends towards $z$, and this limit is then the derivative $f'(z)$ of $f$ at $z$. We say that $f$ is holomorphic if it is derivable at any point in $U$. It is a remarkable feature of complex analysis that such a function is automatically analytic, and therefore in particular indefinitely derivable. In fact, it can be shown that such a function $f:U\to \mathbb C$ is holomorphic if and only if it is analytic! In this case, on any open ball $B_\alpha(z)\subseteq U$ where $f$ is described by $f(w)=\sum_{n=0}^{+\infty} a_n (w-z)^n$, the derivative $f’$ of $f$ is also described by $$f'(w)=\sum_{n=0}^{+\infty} (n+1)a_{n+1} (w-z)^n.$$

2.The complex exponential function

2.1.Extension of \(\exp:\mathbb R\to \mathbb R\) to the complex plane

Let us again take the example of the real exponential function $\exp:\mathbb R\to \mathbb R_+^*$, which is analytical, and which can be expanded into a power series at $0$ for all $x\in \mathbb R$, so that $exp(x)=\sum_{n=0}^{+\infty} \dfrac{x^n}{n!}$. Using the previous notation, this expansion is described by the sequence $(a_n)=\left(\dfrac{1}{n!}\right)$ of real numbers. We can see that $\exp'(x)=\exp(x)$ for any real number $x$, since if we “derive” the series term by term, we obtain $$\begin{eqnarray}\exp'(x)&=&\sum_{n=0}^{+\infty}(n+1)a_{n+1}x^n\\&=&\sum_{n=0}^{+\infty} (n+1)\dfrac{1}{(n+1)! } x^n\\&=&\sum_{n=0}^{+\infty} \dfrac{x^n}{n!},\end{eqnarray}$$ by definition of $(n+1)!$

That being said, the sequence $\left(\dfrac{1}{n!}\right)$ is also a sequence of complex numbers, and by the properties of complex number series, for any such number $z\in\mathbb C$ the series $\sum \dfrac{z^n}{n!}$ converges. This shows that we can define a complex analytic function, again denoted $\exp$, on the whole complex plane by putting $\exp(z)=\sum_{n=0}^{+\infty} \dfrac{z^n}{n!}$. This is the complex exponential function, which is an extension of the real exponential.

2.2. Elementary properties of the complex exponential function

The definition of the complex exponential by a series makes it possible to show, by the rules of calculation on the complex series, that for all complex numbers $z$ and $w$ we have $\exp(z+w)=\exp(z)\times \exp(w)$. In other words, the exponential transforms addition in $\mathbb C$ into multiplication in $\mathbb C^*$. This is referred to as a group homomorphism from the group $(\mathbb C,+)$ into the group $(\mathbb C^,\times)$.

Indeed, if $z=x+iy$ is a complex number, the conjugate $\overline{\exp(z)}$ of the complex number $\exp(z)$ is, by the properties of the limits of complex number sequences, the sum of the series $\sum \dfrac{\overline z^n}{n!}$, i.e. $\sum_{n=0}^{+\infty} \dfrac{\overline z^n}{n!}=\exp(\overline z)$. This means in particular that $|\exp(z)|=\exp(Re(z))=\exp(x)$, since we have $$\begin{eqnarray}|\exp(z)|^2&=&\exp(z).\overline{\exp(z)}=\exp(z).\exp(\overline z)\\&=&\exp(z+\overline z)=\exp(2 Re(z))=\exp(Re(z))^2\end{eqnarray}$$ and we extract the square roots. The modulus $|\exp(iy)|$ of the exponential of a pure imaginary number of the form $iy$ is then $\exp(0)=1$, so in general we have $|\exp(x+iy)|=|\exp(x)||\exp(iy)|=|\exp(x)|\neq 0$, which shows that the complex exponential never becomes zero. It can be shown that the image of the complex exponential is the whole set $\mathbb C^*$. Finally, as a complex analytic function is derivable, and its derivative is given by a series expansion around any point, the function $\exp: \mathbb C\to \mathbb C^*$ is holomorphic, with complex derivative $\exp'(z)=\exp(z)$ at any point $z$, by the same calculation as that used in the real case.

2.3.Determinations of the complex logarithm

What about the logarithm function? From the expansion $\ln(x)=\sum_{n=1}^{+\infty} (-1)^{n+1}\dfrac{(x-1)^n}{n}$ at $x_0=1$, for all $x\in ]0,+\infty[$, of the real logarithm function, we can define a complex logarithm function $\ln: B_1(1)\to \mathbb C$, by putting $\ln(z)=\sum_{n=1}^{+\infty} (-1)^{n+1}\dfrac{(z-1)^n}{n}$ for all $z\in B_1(1)=\{z\in \mathbb C : |z-1|<1\}$. This function is analytic, and by taking $a_0=0$ and $a_n=\dfrac{(-1)^{n+1}}{n}$ for all $n\geq 1$, we can compute its derivative from its expansion. One finds $\ln'(z)=\sum_{n=0}^{+\infty} (n+1)a_{n+1} (z-1)^n=1+\sum_{n=1}^{+\infty}(-1)^n(z-1)^n=\sum_{n=0}^{+\infty} (1-z)^n=\dfrac{1}{1-(1-z)}=\dfrac{1}{z}$, where we recognise the sum of a geometric series with reason $1-z$, for $|1-z|<1$ (in fact, we often start from this second series to establish that the first is an expansion of the logarithm). We can then show that for all $z\in B_1(1)$, we have $\exp\circ \ln(z)=z$, which extends the identity $\exp\circ \ln(x)=x$ for all $x\in ]0,2[$.

However, this ‘complex logarithm’ is not an inverse bijection of the complex exponential, and not by a long shot, since the function $\exp:\mathbb C\to \mathbb C^*$ is not injective. One can show that its kernel (i.e. the set of complex numbers $z$ for which $\exp(z)=1$) is the set $2i\pi\mathbb Z$ of integer multiples of the pure imaginary number $2i\pi$. There are in fact many ‘complex logarithms’ for the complex exponential function – whereas the natural logarithm is the only reciprocal bijection of the real exponential function. Such a function is called a continuous determination of the logarithm, i.e. a continuous mapping $l:X\to \mathbb C$ such that for any $z\in X$, we have $\exp(l(z))=z$, where $X\subseteq \mathbb C^*$ is a complex set not containing the origin. It can in fact be shown that the function $\ln:B_1(1)\to \mathbb C$ defined above extends to a unique continuous determination of the logarithm $\ln : \mathbb C-\mathbb R_-\to \mathbb C$, where $\mathbb R_-=\{x\in \mathbb R : x\leq 0\}$ is the set of negative real numbers. This function is called the principal determination of the complex logarithm.