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.

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle z\)	\(=\)	\(\displaystyle \cos \theta + i \sin \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {\mathrm dz}{\mathrm d\theta}\)	\(=\)	\(\displaystyle -\sin \theta + i\cos \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Derivative of Sine Function, Derivative of Cosine Function, Linear Combination of Derivatives
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle i^2\sin \theta + i\cos \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	$i^2 = -1$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle i\left(i\sin\theta + \cos \theta\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle iz\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\Box$

Step 2

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle y\)	\(=\)	\(\displaystyle e^{i\theta}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {\mathrm dy}{\mathrm d\theta}\)	\(=\)	\(\displaystyle ie^{i\theta}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Derivative of Exponential Function, Chain Rule,Linear Combination of Derivatives
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle iy\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\Box$

Step 3

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left.{z}\right \vert_{\theta=0}\)	\(=\)	\(\displaystyle ie^{i0}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left.{y}\right \vert_{\theta=0}\)	\(=\)	\(\displaystyle i\left(i\sin0 + \cos 0\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

But a specific solution to a differential equation is unique.

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

$\blacksquare$

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {\mathrm d}{\mathrm d \theta} e^{-i \theta} \left({\cos \theta + i \sin \theta}\right)\)	\(=\)	\(\displaystyle e^{-i\theta} \left({-\sin \theta + i \cos \theta}\right) + \left({-i e^{-i\theta} }\right) \left({\cos \theta + i \sin \theta}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Product Rule
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e^{-i \theta} \left({-\sin \theta + i \cos \theta - i \cos \theta - i^2 \sin \theta}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e^{-i \theta} \left({-\sin \theta + i \cos \theta - i \cos \theta + \sin \theta}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e^{-i \theta} \left({0}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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.

$\blacksquare$

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle e^{i \theta}\)	\(=\)	\(\displaystyle \left({1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \frac{\theta^8}{8!} +\cdots}\right) + \left({i\theta - \frac{i\theta^3}{3!} + \frac{i\theta^5}{5!} - \frac{i\theta^7}{7!} + \cdots}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \frac{\theta^8}{8!} +\cdots}\right) + i \left({\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

From the definitions of the sine and cosine functions:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sin \theta\)	\(=\)	\(\displaystyle \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \cos \theta\)	\(=\)	\(\displaystyle 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \cdots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

we obtain the result we want:

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

$\blacksquare$

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

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \arg \left({z_1z_2}\right)\)

\(=\)

\(\displaystyle \arg \left({z_1}\right) + \arg \left({z_2}\right)\)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

Argument of Product is Sum of Arguments

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

• $\operatorname{alog} \left({z_1z_2}\right) = \operatorname{alog} \left({z_1}\right) + \operatorname{alog} \left({z_2}\right)$, and
• $\operatorname{alog} \left({x}\right) =\log{x}$

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \operatorname{alog} \left({z_1z_2}\right)\)	\(=\)	\(\displaystyle a\arg\left({z_1z_2}\right) + \log\left\vert {z_1z_2} \right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a\left[\arg\left({z_1}\right)+\arg\left({z_1}\right)\right] + \log{\left(\left\vert {z_1} \right\vert \left\vert {z_2} \right\vert\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a\left[\arg\left({z_1}\right)+\arg\left({z_1}\right)\right] + \log{\left\vert {z_1} \right\vert} + \log{\left\vert {z_2} \right\vert}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a \arg\left({z_1}\right) + \log{\left\vert {z_1} \right\vert} +a \arg\left({z_2}\right) + \log{\left\vert {z_2} \right\vert}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \operatorname{alog} \left({z_1}\right) + \operatorname{alog} \left({z_1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

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

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 $\operatorname{alog}$:

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

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac{\mathrm{d} }{\mathrm{d}\theta} (a \theta)\)	\(=\)	\(\displaystyle \frac{\mathrm{d} }{\mathrm{d}\theta} \left[\operatorname{alog} \left({\cos\theta+i\sin\theta}\right)\right]\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle a\)	\(=\)	\(\displaystyle \frac{\mathrm{d} \left({\cos\theta+i\sin\theta}\right) }{\mathrm{d}\theta} \frac{\mathrm{d} [\operatorname{alog} \left({\cos\theta+i\sin\theta}\right)] }{\mathrm{d}\left({\cos\theta+i\sin\theta}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Which is the chain rule
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle a\)	\(=\)	\(\displaystyle (-\sin\theta+i\cos\theta) \frac{1}{\cos\theta+i\sin\theta}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from our assumption that $\frac{\mathrm{d}(\operatorname{alog} {z}) }{\mathrm{d}z} = \frac{1}{z}$

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.

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:

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

$\blacksquare$

Source of Name

This entry was named for Leonhard Paul Euler.

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $7.16$
• For a video presentation of the contents of this page, visit the Khan Academy.