# Euler's Formula/Real Domain/Proof 3

## Theorem

Let $\theta \in \R$ be a real number.

Then:

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

## Proof

It follows from Argument of Product equals Sum of Arguments that the $\arg \left({z}\right)$ function for all $z \in \C$ satisfies the relationship:

$\arg \left({z_1 z_2}\right) = \arg \left({z_1}\right) + \arg \left({z_2}\right)$

which means that $\arg \left({z}\right)$ is a kind of logarithm, in the sense that it satisfies the fundamental property of logarithms:

$\log x y = \log x + \log y$

Notice that $\arg \left({z}\right)$ can not be considered a generalization to complex values of the ordinary $\log$ function for real values, since for $x \in \R$, we have:

$0 = \arg \left({x}\right) \ne \log x$

If we do wish to generalize the $\log$ function to complex values, we can use $\arg \left({z}\right)$ to define a set of functions:

$\operatorname{alog} \left({z}\right) = a \arg \left({z}\right) + \log \left\vert{z}\right\vert$

for any $a \in \C$, where $\left\vert{z}\right\vert$ is the modulus of $z$.

All functions satisfy the fundamental property of logarithms and also coincide with the $\log$ function for all real values.

This is established in the following lemma.

### Lemma 1

For all $a,z \in \C$, define the (complex valued) function $\operatorname{alog}$ as:

$\operatorname{alog} \left({z}\right) = a \arg \left({z}\right) + \log \left\vert{z}\right\vert$

then, for any $z_1, z_2 \in \C$ and $x \in \R$:

$\operatorname{alog} \left({z_1 z_2}\right) = \operatorname{alog} \left({z_1}\right) + \operatorname{alog} \left({z_2}\right)$

and:

$\operatorname{alog} \left({x}\right) = \log x$

This means that our (complex valued) $\operatorname{alog}$ functions can genuinely be considered generalizations of the (real valued) $\log$ function.

### Proof of Lemma 1

Let $z_1, z_2$ be any two complex numbers, straightforward substitution on the definition of $\operatorname{alog}$ yields:

 $\displaystyle \operatorname{alog} \left({z_1 z_2}\right)$ $=$ $\displaystyle a \arg \left({z_1 z_2}\right) + \log \left\vert {z_1 z_2} \right\vert$ $\displaystyle$ $=$ $\displaystyle a \left({\arg \left({z_1}\right) + \arg \left({z_1}\right)}\right) + \log \left({\left\vert {z_1} \right\vert \left\vert {z_2} \right\vert}\right)$ $\displaystyle$ $=$ $\displaystyle a \left({\arg \left({z_1}\right) + \arg \left({z_1}\right)}\right) + \log \left\vert {z_1} \right\vert + \log \left\vert {z_2} \right\vert$ $\displaystyle$ $=$ $\displaystyle a \arg \left({z_1}\right) + \log \left\vert {z_1} \right\vert + a \arg \left({z_2}\right) + \log \left\vert {z_2} \right\vert$ $\displaystyle$ $=$ $\displaystyle \operatorname{alog} \left({z_1}\right) + \operatorname{alog} \left({z_1}\right)$

Second part of our lemma is even more straightforward since for $x \in \R$, we have:

$\arg \left({x}\right) = 0$

Then:

 $\displaystyle \operatorname{alog} \left({x}\right)$ $=$ $\displaystyle a \arg \left({x}\right) + \log \left\vert{x}\right\vert$ $\displaystyle$ $=$ $\displaystyle \log x$

which concludes the proof of Lemma 1.

$\Box$

We're left with an infinitude of possible generalizations of the $\log$ function, namely one for each choice of $a$ in our definition of $\operatorname{alog}$.

The following lemma proves that there's a value for $a$ that guarantees our definition of $\operatorname{alog}$ satisfies the much desirable property of $\log$:

$\dfrac{\mathrm d \log x} {\mathrm d x} = \dfrac 1 x$

### Lemma 2

Let $\operatorname{alog} \left({z}\right) = a \arg \left({z}\right) + \log \left|{z}\right|$.

Then if:

$\dfrac {\mathrm d \left({\operatorname{alog} z}\right)} {\mathrm d z} = \dfrac 1 z$

we must have:

$a = i$

### Proof of Lemma 2

Let $z \in \C$ be such that:

$\left\vert{z}\right\vert = 1$

and:

$\arg \left({z}\right) = \theta$

Then:

$z = \cos \theta + i \sin \theta$

Plugging those values in our definition of $\operatorname{alog}$:

 $\displaystyle \operatorname{alog} \left({z}\right)$ $=$ $\displaystyle a \arg \left({\cos \theta + i \sin \theta}\right) + \log \left\vert{z}\right\vert$ $\displaystyle$ $=$ $\displaystyle a \theta + \log 1 = a \theta$

We now have:

$a \theta = \operatorname{alog} \left({\cos \theta + i \sin \theta}\right)$

Taking the derivative with respect to $\theta$ on both sides, we have

 $\displaystyle \frac{\mathrm d}{\mathrm d \theta} (a \theta)$ $=$ $\displaystyle \frac{\mathrm d}{\mathrm d \theta} \left({\operatorname{alog} \left({\cos \theta + i \sin \theta}\right)}\right)$ $\displaystyle a$ $=$ $\displaystyle \dfrac {\mathrm d \left({\cos \theta + i \sin \theta}\right)} {\mathrm d \theta} \dfrac {\mathrm d \left({\operatorname{alog} \left({\cos \theta + i \sin \theta}\right)}\right)} {\mathrm d \left({\cos \theta + i \sin \theta}\right)}$ Chain Rule $\displaystyle a$ $=$ $\displaystyle \left({-\sin \theta + i \cos \theta}\right) \frac 1 {\cos \theta + i \sin \theta}$ from our assumption that $\dfrac {\mathrm d \left({\operatorname{alog} z}\right)} {\mathrm d z} = \dfrac 1 z$

This last equation is true regardless of the value of $\theta$.

In particular, for $\theta = 0$, we must have:

$a = i$

which proves the lemma.

$\Box$

We now have established there is one function which truly deserves to be called the logarithm of complex numbers, defined as:

$\log \left({z}\right) = i \arg \left({z}\right) + \log \left\vert{z}\right\vert$

Since for any $z, z_1, z_2 \in \C, x \in \R$ it satisfies:

$\log \left({z_1 z_2}\right) = \log \left({z_1}\right) + \log \left({z_2}\right)$
$\log \left({x}\right) = \log x$
$\dfrac {\mathrm d \left({\log \left({z}\right)}\right)} {\mathrm d z} = \dfrac 1 z$

Let its inverse function be referred to as the exponential of complex numbers, denoted as $e^z$.

If we write $z$ in its polar form:

$z = \left\vert{z}\right\vert \left({\cos \theta + i \sin \theta}\right)$

we have that:

$e^{i \theta + \log \left\vert{z}\right\vert} = \left\vert{z}\right\vert \left({\cos \theta + i \sin \theta}\right)$

Consider this equation for any number $z$ such that $\left\vert{z}\right\vert = 1$.

Then:

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

$\blacksquare$