log and e

 The mathematical constant $e$, sometimes called Euler's number or Napier's constant, is an irrational number that appears everywhere in mathematics and other sciences and has a value of $e=2.71828\cdots$. Now we see how this constant is defined and how it is related to complex numbers. We assume you know some calculus.

Definition (Natural logarithm)

The natural logarithm function $\log :(0,\mathrm{\infty })\to \mathbb{R}$ is defined by

$$\log x={\int }_1^x\frac{1}{t}dt.$$

From the definition, we can immediately derive a few important properties of $\log$.

1. $\log (1)=0$.
2. $\log$ is a strictly increasing function as $1/t>0$ for all $t>0$. This means, for any $x_1,x_2\in (0,\mathrm{\infty })$, $$x_1<x_2⟹\log x_1<\log x_2.$$
3. $\log$ is a continuous function. This means that for any $a\in (0,\mathrm{\infty })$, we have $$\underset{x\to a}{lim}\log x=\log a.$$

Lemma (The logarithm of a product is the sum of logarithms)

For $y_1,y_2>0,y_1,y_2\in \mathbb{R}$, we have

$$\log (y_1{y}_2)=\log y_1+\log y_2.$$

Proof. 

$$\begin{array}{rcl}\log (y_1{y}_2)& =& {\int }_1^{y_1{y}_2}\frac{1}{x}dx\\ & =& {\int }_1^{y_1}\frac{1}{x}dx+{\int }_{y_1}^{y_1{y}_2}\frac{1}{x}dx\\ & =& \log y_1+{\int }_1^{y_2}\frac{1}{v}dv\end{array}$$

where we changed the variables using $v=x/y_1$ (and hence, $dv=dx/y_1$; $x=y_1\mapsto v=1$; $x=y_1{y}_2\mapsto v=y_2$). But the name of the variable doesn't matter, so we have

$$\log (y_1{y}_2)=\log (y_1)+\log (y_2).$$

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Corollary

For all $x\in (0,\mathrm{\infty })$ and $n\in \mathbb{N}$, we have

$$\log (x^n)=n\log x.$$

Proof. Exercise. (Use mathematical induction.) ■

From this corollary, we have, in particular,

$$\log (2^m)=m\log 2.$$

Since $\log 1=0$ and $\log$ is a strictly increasing function, it follows that $\log 2>0$. Thus, as $m$ increases, $\log (2^m)$ can take arbitrarily large values.

Corollary

For any $x>0$, we have

$$\log (1/x)=-\log x.$$

Proof. Using the above lemma and the property of $\log$, we have

$$0=\log (1)=\log (x\cdot (1/x))=\log x+\log (1/x)$$

from which the desired result follows. ■

If $x>1$, then $1/x<1$. As $x$ increases, $\log (1/x)$ can take arbitrarily large negative values.

Let's summarize the properties of $\log :(0,\mathrm{\infty })\to \mathbb{R}$.

1. It is a strictly increasing function. Hence it is injective.
2. It takes all real values. Hence surjective (Note the domain and codomain).

Thus $\log$ is bijective and hence has an inverse which we call $\exp :\mathbb{R}\to (0,\mathrm{\infty })$.

Definition (Exponential $\exp$)

The exponential function $\exp :\mathbb{R}\to (0,\mathrm{\infty })$ is defined as the inverse of the natural logarithm function $\log :(0,\mathrm{\infty })\to \mathbb{R}$.

Note, in particular, $\exp (0)=1$ as $\log (1)=0$.

Definition ($e$)

The number $e$ is defined by

$$e=\exp (1).$$

In the following, we show that $\exp$ is indeed the exponential function with base $e$, that is, $\exp (x)=e^x$ for all $x\in \mathbb{R}$.

Lemma (The exponential of a sum is the product of exponentials)

For any $x_1,x_2\in \mathbb{R}$,

$$\exp (x_1+x_2)=\exp (x_1)\exp (x_2).$$

Proof. Let $y_1=\exp (x_1)$ and $y_2=\exp (x_2)$. Then, $x_1=\log y_1$ and $x_2=\log y_2$. By the property of the natural logarithm, we have

$$\log (y_1{y}_2)=\log y_1+\log y_2=x_1+x_2.$$

Thus, 

$$y_1{y}_2=\exp (x_1+x_2).$$

Substituting the definitions of $y_1$ and $y_2$, we conclude that

$$\exp (x_1+x_2)=\exp (x_1)\exp (x_2)$$

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It follows that, for any $n\in \mathbb{N}$,

$$\exp (n)=\{\exp (1){\}}^n=e^n.$$

Also, for any $n\in \mathbb{N}$,

$$\exp (-n)\exp (n)=\exp (0)=1$$

so that

$$\exp (-n)=e^{-n}.$$

Thus, for any $z\in \mathbb{Z}$, we have

$$\exp (z)=\{\exp (1){\}}^z=e^z.$$

Next, note that $\exp (1/n)$ is real and positive, and

$$(\exp (1/n){)}^n=\exp (n/n)=\exp (1)=e.$$

Thus $\exp (1/n)$ is the unique real $n$-th root of $e$: 

$$\exp (1/n)=e^{1/n}.$$

Next, consider the rational $m/n$ where $m\in \mathbb{Z},n\in \mathbb{N}$.

$$\exp (m/n)=(\exp (1/n){)}^m=(e^{1/n}{)}^m=e^{m/n}.$$

So now we know that

$$\exp (x)=e^x,\mathrm{\forall }x\in \mathbb{Q}.$$

However, $\exp (x)$ and $e^x$ are both continuous and all real numbers can be approximated by rational numbers, so we have

$$\exp (x)=e^x,\mathrm{\forall }x\in \mathbb{R}.$$

Now we know that $\exp (x)$ is the exponential function with base $e$, namely, $e^x$. The next lemma is one of the conspicuous properties of $\exp (x)$.

Lemma

$$\frac{d}{dx}{e}^x=e^x.$$

Proof. Let $y=\exp (x)$. Then $x=\log (y)$. Differentiating, we have

$$1=\frac{1}{y}\frac{dy}{dx},$$

so

$$\frac{dy}{dx}=y.$$

That is,

$$\frac{d}{dx}\exp (x)=\exp (x).$$

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Next, we would like to define $e^z$ for $z\in \mathbb{C}$.

Definition ($e^z$ on the complex domain)

Let $z=u+i\theta$ where $u,\theta \in \mathbb{R}$. We define $e^z$ by

$$e^z=e^u(\cos \theta +i\sin \theta ).$$

Remark. Don't get confused. ``$e^u$'' with $u\in \mathbb{R}$ means the function $\exp (u):\mathbb{R}\to \mathbb{R}$ as originally defined above, whereas ``$e^z:\mathbb{C}\to \mathbb{C}$'' is a new function being defined with a different domain and codomain. □

To see that this definition is consistent with the previous definition of $e^x$ for $x\in \mathbb{R}$, let $\theta =0$ so the ``complex number'' is purely real. Since $\cos 0=1$ and $\sin 0=0$, the new definition matches the old definition.

Example. The famous equality

$$e^{i\pi }=-1$$

is called Euler's formula. □

We can further show that $e^z$ on the complex domain behaves in the same manner as $e^x$ on the real domain.

Lemma

Let $z_1,z_2\in \mathbb{C}$ and $k\in \mathbb{Z}$. The following equations hold.

1. $e^{z_1}{e}^{z_2}=e^{z_1+z_2}$.
2. $(e^{z_1}{)}^k=e^{k{z}_1}$.

Proof. We may assume that $z_1=u_1+i{\theta }_1$ and $z_2=u_2+i{\theta }_2$ for some $u_1,u_2,{\theta }_1,{\theta }_2\in \mathbb{R}$.

1. $$\begin{array}{rcl}{e}^{z_1}{e}^{z_2}& =& e^{u_1}(\cos {\theta }_1+i\sin {\theta }_1)e^{u_2}(\cos {\theta }_2+i\sin {\theta }_2)\\ & =& e^{u_1}{e}^{u_2}(\cos {\theta }_1+i\sin {\theta }_1)(\cos {\theta }_2+i\sin {\theta }_2)\\ & =& e^{u_1+u_2}(\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2))\\ & =& e^{z_1+z_2}\end{array}$$ as $z_1+z_2=(u_1+u_2)+i({\theta }_1+{\theta }_2)$.
2. If $k>0$, this trivially follows from De Moivre's theorem. If $k=0$, it is trivial. For $k<0$, note that $(e^z{)}^{-1}=e^{-z}$ for all $z\in \mathbb{C}$ (use Part 1 to see this); Let $k^{\prime }=-k$ so $k^{\prime }>0$. $$\begin{array}{rcl}(e^{z_1}{)}^k& =& (e^{z_1}{)}^{-1\cdot k^{\prime }}\\ & =& \underset{k^{\prime }\text{ times}}{\underbrace{(e^{z_1}{)}^{-1}\cdots (e^{z_1}{)}^{-1}}}\\ & =& e^{-z_1}\cdots e^{-z_1}\\ & =& e^{-k^{\prime }{z}_1}=e^{k{z}_1}.\end{array}$$

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Polar form revisited

We know that, for $\theta \in \mathbb{R}$,

$$e^{i\theta }=\cos \theta +i\sin \theta .$$

So, the polar form of a non-zero complex number

$$r(\cos \theta +i\sin \theta )$$

is expressed simply as

$$r{e}^{i\theta }.$$

That is, any complex number $z\in \mathbb{C}$ can be represented as

$$z=r{e}^{i\theta }$$

where $r=|z|$ and $\theta \in \arg z$.

We have

$$(r{e}^{i\theta }{)}^{-1}=(r^{-1})e^{-i\theta }$$

and

$$r{e}^{i\theta }\cdot s{e}^{i\psi }=(rs)e^{i(\theta +\psi )}.$$

It follows that

$$(r{e}^{i\theta })/(s{e}^{i\psi })=(r{s}^{-1})e^{i(\theta -\psi )}.$$

Since $e^{i\theta }=\cos \theta +i\sin \theta$ and $e^{-i\theta }=\cos \theta -i\sin \theta$, we have

$$\begin{array}{rcl}\cos \theta & =& \frac{e^{i\theta }+e^{-i\theta }}{2},\\ \sin \theta & =& \frac{e^{i\theta }-e^{-i\theta }}{2i}.\end{array}$$