3.3 The Exponential Function

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The exponential function is the power series \[\exp(z) = \sum_{n=0}^\infty \frac{z^n}{n!} = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots\] which has infinite radius of convergence i.e. the series converges absolutely for every $z \in \C$. We are familiar with the exponential function $e^x$ of a real variable, which has the property that $(e^x)' = e^x$. The complex exponential has the same property.

Lemma

One has $\exp'(z) = \exp(z)$ for all $z \in \C$.

Proof:

All the hard work was in the previous section: we proved that power series are differentiable inside their radius of convergence, and that the derivative can be calculated term-by-term as if the series were a finite sum. So \[\exp' (z) = (1)' + \sum_{n=1}^\infty \left( \frac{z^n}{n!} \right)' = \sum_{n=1}^\infty \frac{nz^{n-1}}{n!} = \sum_{n=1}^\infty \frac{z^{n-1}}{(n-1)!} = \exp(z)\] as claimed. $\square$

The additive property $e^{x+y} = e^x e^y$ also survives the passage to complex analysis. To prove it we need the following lemma.

Lemma

If $f : \C \to \C$ is holomorphic and $f'(z) = 0$ for all $z \in \C$ then $f$ is constant.

Proof:

Writing $f(x+iy) = u(x,y) + i v(x,y)$ the Cauchy-Riemann equations tell us that \[f'(z) = \partial_1 u + i \partial_1 v = \partial_2 v - i \partial_2 u\] so $\partial_1 u$ and $\partial_2 u$ are zero everywhere. It follows that $u$ is constant on every horizontal line and on every vertical line. Thus $u(x,y) = u(x,0) = u(0,0)$ for all $x,y \in \R$. A similar argument proves that $v$ is constant on $\C$ and therefore $f$ is as well. $\square$

Lemma

We have $\exp(z+w) = \exp(z) \exp(w)$ for all $z,w \in \C$.

Proof:

Fix $c \in \C$ and define $f(z) = \exp(z) \exp(c-z)$. We calculate \[f'(z) = \exp(z) \exp(c-z) - \exp(z)\exp(c-z) = 0\] for all $z \in \C$ using the product rule and the chain rule. Thus $f' = 0$ on all of $\C$. By the previous lemma $f$ is constant i.e. $f(z) = f(0) = \exp(c)$ for all $z,c \in \C$. Taking $c = z+w$ gives the conclusion. $\square$

If we take $w = -z$ in the above result we get \[1 = \exp(0) = \exp(z) \exp(-z)\] which implies $\exp(-z)$ is the multiplicative inverse of $\exp(z)$. In particular $\exp(z)$ is non-zero for all $z \in \C$. As we will see later, zero is the only complex number that is not in the range of the exponential function.

Define $e = \exp(1)$. Thus $e$ is the number \[e = \sum_{n=0}^\infty \frac{1}{n!} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \approx 2.7182\cdots\] and by induction $e^n = \exp(1)^n = \exp(n)$ for all $n \in \N$. More generally, for any $n,m \in \N$ we have \[\exp(m/n)^n = \exp(m) = e^m\] so $\exp(m/n) = e^{m/n}$. Continuity then implies that $\exp(x) = e^x$ for all $x > 0$.

If we take $z = x$, $w = iy$ then \[\exp(z) = \exp(x) \exp(iy) = e^x \exp(iy)\] and we will have a good understanding of the exponential function if we can understand $\exp(iy)$ for $y \in \R$. Remarkably, the exponential function at imaginary numbers is closely related to trigonometric functions!