4.2 The Exponential Function

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We introduced the exponential function in section on power series. Recall that $$\mathsf{exp}(z)=\sum _{n=0}^{\mathrm{\infty }}\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$$ has infinite radius of convergence i.e. the series converges absolutely for every $z\in \mathbb{C}$. We are familiar with the exponential function $e^x$ of a real variable, which has the property that $(e^x{)}^{\prime }=e^x$. The complex exponential has the same property.

Lemma

One has ${\mathsf{exp}}^{\prime }(z)=\mathsf{exp}(z)$ for all $z\in \mathbb{C}$.

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 $${\mathsf{exp}}^{\prime }(z)=(1{)}^{\prime }+\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{z^n}{n!}\right)}^{\prime }=\sum _{n=1}^{\mathrm{\infty }}\frac{n{z}^{n-1}}{n!}=\sum _{n=1}^{\mathrm{\infty }}\frac{z^{n-1}}{(n-1)!}=\mathsf{exp}(z)$$ as claimed. $◻$

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 $f:\mathbb{C}\to \mathbb{C}$ is holomorphic and $f^{\prime }(z)=0$ for all $z\in \mathbb{C}$ then $f$ is constant.

Proof:

Writing $f(x+iy)=u(x,y)+iv(x,y)$ the Cauchy-Riemann equations tell us that $$f^{\prime }(z)={\partial }_{1}u+i{\partial }_{1}v={\partial }_{2}v-i{\partial }_{2}u$$ 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 $x,y\in \mathbb{R}$. A similar argument proves that $v$ is constant on $\mathbb{C}$ and therefore $f$ is as well. $◻$

Lemma

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

Proof:

Fix $c\in \mathbb{C}$ and define $f(z)=\mathsf{exp}(z)\mathsf{exp}(c-z)$. We calculate $$f^{\prime }(z)=\mathsf{exp}(z)\mathsf{exp}(c-z)-\mathsf{exp}(z)\mathsf{exp}(c-z)=0$$ for all $z\in \mathbb{C}$ using the product rule and the chain rule. Thus $f^{\prime }=0$ on all of $\mathbb{C}$. By the previous lemma $f$ is constant i.e. $f(z)=f(0)=\mathsf{exp}(c)$ for all $z,c\in \mathbb{C}$. Taking $c=z+w$ gives the conclusion. $◻$

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

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

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