Euler's Formula

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Theorem

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



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


Thus the complex exponential function is defined in terms of standard trigonometric functions.


Corollary

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


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.

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

$\Box$


Step 2

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

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

$\Box$


Step 3

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

\(\displaystyle \left.{y}\right \vert_{\theta \mathop = 0}\) \(=\) \(\displaystyle e^{0i}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 1\) $\quad$ $\quad$
\(\displaystyle \left.{z}\right \vert_{\theta \mathop = 0}\) \(=\) \(\displaystyle \cos 0 + i \sin 0\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 1\) $\quad$ $\quad$

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

But a specific solution to a differential equation is unique.


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

$\blacksquare$


Proof 2

This:

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

is logically equivalent to this:

$\dfrac{\cos \theta + i \sin \theta} {e^{i \theta}} = 1$

for every $\theta$.

Note that the left expression is nowhere undefined.


Taking the derivative of this:

\(\displaystyle \dfrac {\mathrm d}{\mathrm d \theta} e^{-i \theta} \left({\cos \theta + i \sin \theta}\right)\) \(=\) \(\displaystyle e^{-i\theta} \left({-\sin \theta + i \cos \theta}\right) + \left({-i e^{-i\theta} }\right) \left({\cos \theta + i \sin \theta}\right)\) $\quad$ Product Rule and Derivative of Exponential Function $\quad$
\(\displaystyle \) \(=\) \(\displaystyle e^{-i \theta} \left({-\sin \theta + i \cos \theta - i \cos \theta - i^2 \sin \theta}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle e^{-i \theta} \left({-\sin \theta + i \cos \theta - i \cos \theta + \sin \theta}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle e^{-i \theta} \left({0}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 0\) $\quad$ $\quad$


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 $\theta = 0$:

$\dfrac {\cos 0 + i \sin 0} {e^{0 i}} = \dfrac {1 + 0} 1 = 1$

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

Hence the result.

$\blacksquare$


Proof 3

As Sine Function is Absolutely Convergent and Cosine Function is Absolutely Convergent, we have:

\(\displaystyle \cos \theta + i \sin \theta\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \dfrac{\theta^{2 n} }{\left({2 n}\right)! } + i \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \dfrac{\theta^{2 n + 1} }{\left({2 n + 1}\right)! }\) $\quad$ Definition of Complex Cosine Function and Definition of Complex Sine Function $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \left({\left({-1}\right)^n \dfrac{\theta^{2 n} }{\left({2 n}\right)! } + i \left({-1}\right)^n \dfrac{\theta^{2 n + 1} }{\left({2 n + 1}\right)! } }\right)\) $\quad$ by Sum of Absolutely Convergent Series $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \left({ \dfrac{ \left({i \theta}\right)^{2 n} }{\left({2 n}\right)! } + \dfrac{ \left({i \theta}\right)^{2 n + 1} }{\left({2 n + 1}\right)! } }\right)\) $\quad$ as $i^2 = -1$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \dfrac{ \left({i \theta}\right)^n }{n!}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle e^{i \theta}\) $\quad$ Definition of Complex Exponential Function $\quad$

$\blacksquare$


Proof 4

It follows from Argument of Product is 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:

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


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

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

Then:

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

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 $\operatorname{alog}$:

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

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

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

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:

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

$\blacksquare$


Proof 5

Note that the following proof, as written, only holds for real $\theta$.

Define:

$x \left({\theta}\right) = e^{i \theta}$

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

Consider first $\theta \ge 0$.

Taking Laplace transforms:

\(\displaystyle \mathcal L \left\{ {x \left({\theta}\right)}\right\} \left({s}\right)\) \(=\) \(\displaystyle \mathcal L \left\{ {e^{i \theta} }\right\} \left({s}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {s - i}\) $\quad$ Laplace Transform of Exponential $\quad$
\(\displaystyle \mathcal L \left\{ {y \left({\theta}\right)}\right\} \left({s}\right)\) \(=\) \(\displaystyle \mathcal L \left\{ {\cos \theta + i \sin \theta }\right\}\left({s}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \mathcal L \left\{ {\cos \theta} \right\} \left({s}\right) + i \mathcal L \left\{ { \sin \theta}\right\} \left({s}\right)\) $\quad$ Linear Combination of Laplace Transforms $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac s {s^2 + 1} + \frac i {s^2 + 1}\) $\quad$ Laplace Transform of Cosine, Laplace Transform of Sine $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {s + i} {\left({s + i}\right) \left({s - i}\right)}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {s - i}\) $\quad$ $\quad$

So $x$ and $y$ have the same Laplace transform for $\theta \ge 0$.

Now define $\tau = -\theta, \sigma = -s$, and consider $\theta < 0$ so that $\tau > 0$.

Taking Laplace transforms of $x\left({\tau}\right)$ and $y\left({\tau}\right)$:

\(\displaystyle \mathcal L \left\{ {x \left({\tau}\right)}\right\} \left({\sigma}\right)\) \(=\) \(\displaystyle \frac 1 {\sigma - i}\) $\quad$ from above $\quad$
\(\displaystyle \mathcal L \left\{ {y \left({\tau}\right)}\right\} \left({\sigma}\right)\) \(=\) \(\displaystyle \frac 1 {\sigma - i}\) $\quad$ from above $\quad$

So $x\left({\theta}\right)$ and $y\left({\theta}\right)$ have the same Laplace transform for $\theta < 0$.

The result follows from Injectivity of Laplace Transform.

$\blacksquare$


Also known as

Euler's Formula in this and its corollary form are also found referred to as Euler's Identities, but this term is also used for a different result: $e^{i \pi} + 1 = 0$.

It is wise when referring to it by name, therefore, to ensure that the equation itself is also specified.


Source of Name

This entry was named for Leonhard Paul Euler.


Sources