Euler's Formula

Theorem

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

where $e^\cdot$ is the complex exponential function, $\cos$ is cosine, $\sin$ is sine, and $i$ is the imaginary unit.

Thus we define the complex exponential function in terms of standard trigonometric functions.

Direct Proof 1
Consider the differential equation:


 * $D_z f\left({z}\right) = i \cdot f\left({z}\right)$

Step 1
We will prove that $z = \cos \theta + i \sin \theta$ is a solution.

Step 2
We will prove that $y = e^{i\theta}$ is a solution.

Step 3
Consider the initial condition $f\left({0}\right) = 1$.

So $y$ and $z$ are both specific solutions.

But a specific solution to a differential equation is unique.

Therefore $y = z$, that is, $e^{i \theta} = \cos \theta + i \sin \theta$.

Direct Proof 2
This:
 * $e^{i \theta} = \cos \theta + i \sin \theta$

is logically equivalent to this:
 * $\displaystyle \frac{\cos \theta + i \sin \theta} {e^{i \theta} } = 1$

for every $\theta$.

Note that the left expression is nowhere undefined.

Taking the derivative of this:

Thus the expression, as a function of $\theta$, is constant and so yields the same value for every $\theta$.

We know the value at at least one point, that is, when $\theta = 0$:


 * $\displaystyle \frac{\cos 0 + i \sin 0}{e^{i 0}} = \frac {1 + 0} 1 = 1$

Thus it is $1$ for every $\theta$, which verifies the above.

Hence the result.

Direct Proof 3
Use the Taylor Series Expansion for Exponential Function, we have:


 * $\displaystyle e^{i \theta} = 1 + i \theta + \frac {i^2\theta^2} {2!} + \frac {i^3 \theta^3} {3!} + \frac {i^4 \theta^4} {4!} + \frac {i^5\theta^5} {5!} + \frac {i^6 \theta^6} {6!} + \frac {i^7 \theta^7} {7!} + \frac {i^8 \theta^8} {8!} + \cdots$

The equation can be simplified to


 * $\displaystyle e^{i \theta} = 1 + i \theta - \frac {\theta^2} {2!} - \frac {i \theta^3} {3!} + \frac {\theta^4} {4!} + \frac {i \theta^5} {5!} - \frac {\theta^6} {6!} - \frac {i \theta^7}{7!} + \frac{\theta^8}{8!} + \cdots$

Rearranging the above eqn, we obtain

From the definitions of the sine and cosine functions:

we obtain the result we want:
 * $ e^{i \theta} = \cos \theta + i \sin \theta$

Proof from the properties of the $\arg{(z)}$ function
It is a consequence of the definition of complex multiplication that the $\arg \left({z}\right)$ function, for $z \in \C$, satisfies the relationship

Which means that $\arg \left({z}\right)$ is a kind of logarithm, in the sense that it satisfies the fundamental property of logarithms: $\log{xy}=\log{x}+\log{y}$.

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

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
 * $\operatorname{alog} \left({z}\right) = a\arg\left({z}\right) + \log\left|{z}\right|$, for any $a \in \C$, where $|z|$ is the modulus of $z$,

which 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 any $a,z\in\C$, define the (complex valued) function $\operatorname{alog}$ as
 * $\operatorname{alog} \left({z}\right) = a\arg\left({z}\right) + \log\left|{z}\right|$

then, for any $z_1,z_2\in\C$, and $x\in\R$ we have This means that our (complex valued) $\operatorname{alog}$ functions can genuinely be considered generalizations of the (real valued) $\log$ function.
 * $\operatorname{alog} \left({z_1z_2}\right) = \operatorname{alog} \left({z_1}\right) + \operatorname{alog} \left({z_2}\right)$, and
 * $\operatorname{alog} \left({x}\right) =\log{x}$

Proof of Lemma 1
Let $z_1,z_2$ be any two complex numbers, straightforward substitution on the definition of $\operatorname{alog}$ yields:

Second part of our lemma is even more straightforward since for $x\in\R$, we have $\arg\left(x\right)=0$, then

Which concludes the proof of Lemma 1.

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:
 * $\dfrac{\mathrm{d}({\operatorname{alog}{z}})}{\mathrm{d}{z}}=\dfrac{1}{z}$,

we must have
 * $a=i$.

Proof Lemma 2
Let $z\in\C$ be such that $\left\vert z \right\vert = 1$, and $\arg{(z)}=\theta$ then
 * $z=\cos{\theta}+i\sin{\theta}$

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

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

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.

We now have established there is one function which truly deserves to be called the logarithm of complex numbers, defined as:
 * ${\bf log}(z) = i\arg\left({z}\right) + \log\left|{z}\right| $

Since for any $z,z_1,z_2\in\C,x\in\R$ it satisfies:
 * ${\bf log}(z_1z_2) = {\bf log}(z_1)+{\bf log}(z_2)$
 * ${\bf log}(x) = \log{x}$
 * $\dfrac{\mathrm{d}[{\bf log}(z)]} {\mathrm{d}{z}}=\dfrac{1}{z}$

Lets call it's inverse function the exponential of complex numbers, denoted as $e^z$. If we write $z$ in it's polar form $z=|z|(\cos{\theta}+i\sin{\theta})$, we have that
 * $e^{i\theta + \log\left|{z}\right|}=|z|(\cos{\theta}+i\sin{\theta})$

Consider this equation for any number $z$ such that $|z|=1$, then:
 * $e^{i\theta}=\cos{\theta}+i\sin{\theta}$