Euler's Formula/Real Domain

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


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


$e^{i \theta}$ denotes the complex exponential function
$\cos \theta$ denotes the real cosine function
$\sin \theta$ denotes the real sine function
$i$ denotes the imaginary unit.


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

Proof 1

Consider the differential equation:

$D_z \map f z = i \cdot \map f z$

Step 1

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

\(\ds z\) \(=\) \(\ds \cos \theta + i \sin \theta\)
\(\ds \frac {\d z} {\d \theta}\) \(=\) \(\ds -\sin \theta + i \cos \theta\) Derivative of Sine Function, Derivative of Cosine Function, Linear Combination of Derivatives
\(\ds \) \(=\) \(\ds i^2 \sin \theta + i\cos \theta\) $i^2 = -1$
\(\ds \) \(=\) \(\ds i \paren {i \sin \theta + \cos \theta}\)
\(\ds \) \(=\) \(\ds i z\)


Step 2

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

\(\ds y\) \(=\) \(\ds e^{i\theta}\)
\(\ds \frac {\d y} {\d \theta}\) \(=\) \(\ds i e^{i \theta}\) Derivative of Exponential Function, Chain Rule for Derivatives, Linear Combination of Derivatives
\(\ds \) \(=\) \(\ds i y\)


Step 3

Consider the initial condition $\map f 0 = 1$.

\(\ds \bigintlimits y {\theta \mathop = 0} {}\) \(=\) \(\ds e^{0 i}\)
\(\ds \) \(=\) \(\ds 1\)
\(\ds \bigintlimits z {\theta \mathop = 0} {}\) \(=\) \(\ds \cos 0 + i \sin 0\)
\(\ds \) \(=\) \(\ds 1\)

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

But a particular solution to a differential equation is unique.

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


Proof 2


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

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:

\(\ds \dfrac \d {\d \theta} e^{-i \theta} \paren {\cos \theta + i \sin \theta}\) \(=\) \(\ds e^{-i \theta} \paren {-\sin \theta + i \cos \theta} + \paren {-i e^{-i \theta} } \paren {\cos \theta + i \sin \theta}\) Product Rule and Derivative of Exponential Function
\(\ds \) \(=\) \(\ds e^{-i \theta} \paren {-\sin \theta + i \cos \theta - i \cos \theta - i^2 \sin \theta}\)
\(\ds \) \(=\) \(\ds e^{-i \theta} \paren {-\sin \theta + i \cos \theta - i \cos \theta + \sin \theta}\)
\(\ds \) \(=\) \(\ds e^{-i \theta} \paren 0\)
\(\ds \) \(=\) \(\ds 0\)

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.


Proof 3

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 $x \in \R$:

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


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

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:

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

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

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


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

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:

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

we must have:

$a = i$

Proof of Lemma 2

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

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


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


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

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

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

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

\(\ds \frac{\mathrm d}{\mathrm d \theta} (a \theta)\) \(=\) \(\ds \frac{\mathrm d}{\mathrm d \theta} \left({\operatorname{alog} \left({\cos \theta + i \sin \theta}\right)}\right)\)
\(\ds a\) \(=\) \(\ds \dfrac {\mathrm d \left({\cos \theta + i \sin \theta}\right)} {\mathrm d \theta} \dfrac {\mathrm d \left({\operatorname{alog} \left({\cos \theta + i \sin \theta}\right)}\right)} {\mathrm d \left({\cos \theta + i \sin \theta}\right)}\) Chain Rule for Derivatives
\(\ds a\) \(=\) \(\ds \left({-\sin \theta + i \cos \theta}\right) \frac 1 {\cos \theta + i \sin \theta}\) 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.


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


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


Proof 4

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


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

Consider first $\theta \ge 0$.

Taking Laplace transforms:

\(\ds \map {\laptrans {\map x \theta} } s\) \(=\) \(\ds \map {\laptrans {e^{i \theta} } } s\)
\(\ds \) \(=\) \(\ds \frac 1 {s - i}\) Laplace Transform of Exponential
\(\ds \map {\laptrans {\map y \theta} } s\) \(=\) \(\ds \map {\laptrans {\cos \theta + i \sin \theta} } s\)
\(\ds \) \(=\) \(\ds \map {\laptrans {\cos \theta} } s + i \, \map {\laptrans {\sin \theta} } s\) Linear Combination of Laplace Transforms
\(\ds \) \(=\) \(\ds \frac s {s^2 + 1} + \frac i {s^2 + 1}\) Laplace Transform of Cosine, Laplace Transform of Sine
\(\ds \) \(=\) \(\ds \frac {s + i} {\paren {s + i} \paren {s - i} }\)
\(\ds \) \(=\) \(\ds \frac 1 {s - i}\)

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 $\map x \tau$ and $\map y \tau$:

\(\ds \map {\laptrans {\map x \tau} } \sigma\) \(=\) \(\ds \frac 1 {\sigma - i}\) from above
\(\ds \map {\laptrans {\map y \tau} } \sigma\) \(=\) \(\ds \frac 1 {\sigma - i}\) from above

So $\map x \theta$ and $\map y \theta$ have the same Laplace transforms for $\theta < 0$.

The result follows from Injectivity of Laplace Transform.


Proof 5

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

\(\ds \cos \theta + i \sin \theta\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {\theta^{2 n} } {\paren {2 n}!} + i \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {\theta^{2 n + 1} } {\paren {2 n + 1}!}\) Definition of Complex Cosine Function and Definition of Complex Sine Function
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {\paren {-1}^n \dfrac {\theta^{2 n} } {\paren {2 n}!} + i \paren {-1}^n \dfrac {\theta^{2 n + 1} } {\paren {2 n + 1}!} }\) Sum of Absolutely Convergent Series
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {\dfrac {\paren {i \theta}^{2 n} } {\paren {2 n}!} + \dfrac {\paren {i \theta}^{2 n + 1} } {\paren {2 n + 1}!} }\) Definition of Imaginary Unit
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \dfrac {\paren {i \theta}^n} {n!}\)
\(\ds \) \(=\) \(\ds e^{i \theta}\) Definition of Complex Exponential Function



Example: $e^{i \pi / 4}$

$e^{i \pi / 4} = \dfrac {1 + i} {\sqrt 2}$

Example: $e^{i \pi / 2}$

$e^{i \pi / 2} = i$

Example: $e^{-i \pi / 2}$

$e^{-i \pi / 2} = -i$

Example: $e^{i \pi}$

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

Example: $e^{2 i \pi}$

$e^{2 i \pi} = 1$

Example: $e^{2 k i \pi}$

$\forall k \in \Z: e^{2 k i \pi} = 1$

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 the specific example:

$e^{i \pi} + 1 = 0$

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.