Euler's Formula | Brilliant Math & Science Wiki (2024)

In complex analysis, Euler's formula provides a fundamental bridge between the exponential function and the trigonometric functions. For complex numbers \( x \), Euler's formula says that

\[ e^{ix} = \cos{x} + i \sin{x}. \]

In addition to its role as a fundamental mathematical result, Euler's formula has numerous applications in physics and engineering.

A straightforward proof of Euler's formula can be had simply by equating the power series representations of the terms in the formula:

\[ \cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \]

and

\[ \sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots, \]

so

\[ \begin{align*}\cos{x} + i \sin{x} &= 1 + ix - \frac{x^2}{2!} - i \frac{x^3}{3!} + \frac{x^4}{4!} - \cdots \\&= 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \cdots \\&= e^{ix}.\end{align*} \]

Euler's Formula

Suppose \( x \) is complex. Then

\[ e^{ix} = \cos{x} + i \sin{x}. \]

Compute \( e^{i \pi} \).

We have

\[ e^{i \pi} = \cos{\pi} + i \sin{\pi} = -1,\]

which leads to the very famous Euler's identity: \(e^{i \pi} + 1 = 0.\ _\square\)

Compute \( i^i \).

Recall that \(\forall k \in \mathbb{N}\) we have that

\[ \begin{align*}e^{i (\pi / 2+2\pi k)} &= i \\ \Rightarrow \left(e^{i (\pi / 2+2\pi k)}\right)^i &= i^i.\end{align*} \]

Therefore,

\[ i^i = e^{i^2 (\pi / 2+2\pi k)} = e^{- \pi / 2-2\pi k}.\ _\square \]

Note: This means that \(i^i\) is not a well defined (unique) quantity. To remedy this, one needs to specify a branch cut. For example, we can define the argument of \(e^{i \theta}\) to be defined for \(\theta \in [0, 2\pi)\), in which case we have that \(i^i = e^{-\pi / 2}\). That is, this forces \(k = 0\). Of course, different branch cut can be chosen yielding different values for \(k\).

Euler's formula allows for any complex number \( x \) to be represented as \( e^{ix} \), which sits on a unit circle with real and imaginary components \( \cos{x} \) and \( \sin{x} \), respectively. Various operations (such as finding the roots of unity) can then be viewed as rotations along the unit circle.

One immediate application of Euler's formula is to extend the definition of the trigonometric functions to allow for arguments that extend the range of the functions beyond what is allowed under the real numbers.

A couple useful results to have at hand are the facts that

\[ e^{-ix} = \cos{x} - i \sin{x}, \]

so

\[ e^{ix} + e^{-ix} = 2 \cos{x}. \]

It follows that

\[ \cos{x} = \frac{e^{ix} + e^{-ix}}{2}, \]

and similarly

\[ \sin{x} = \frac{e^{ix} - e^{-ix}}{2i} \]

and

\[ \tan{x} = \frac{e^{ix} - e^{-ix}}{i(e^{ix} + e^{-ix})}. \]

Solve \( \cos{x} = 2 \) in the complex numbers.

We first note that if \(x = x_0 \) is a solution, then so is \(x = 2\pi k \pm x_0 \) for any integer \(k\). This is because \(\cos x\) is an even function with a fundamental period of \(2\pi \).

Taking \( \cos{x} = \dfrac{e^{ix} + e^{-ix}}{2} \) yields

\[ \begin{align*} e^{ix}+e^{-ix} &= 4 \\\left(e^{ix}\right)^2-4e^{ix}+1 &= 0 \\e^{ix} &= 2\pm \sqrt{3} \\\Rightarrow x &= \dfrac1i \ln \left (2 \pm \sqrt 3\right) \\& =- i \ln \left (2 \pm \sqrt 3\right).\end{align*} \]

Hence, \(x = 2 \pi k \pm i \ln \left ( 2\pm \sqrt{3} \right), 2 \pi k \mp i \ln \left ( 2\pm \sqrt{3} \right) \) for any integer \(k\). \( \ _\square\)

Euler’s formula also allows for the derivation of several trigonometric identities quite easily. Starting with

\[ e^{i(x \pm y)} = \cos(x \pm y) + i \sin(x \pm y), \]

one finds

\[\begin{align*}e^{i(x \pm y)} &= e^{ix} e^{\pm iy} \\&= (\cos{x} + i\sin{x})(\cos{y} \pm i \sin{y}) \\&= \cos{x} \cos{y} \mp \sin{x} \sin{y} + i(\sin{x} \cos{y} \pm \cos{x} \sin{y}).\end{align*} \]

Equating the real and imaginary parts, respectively, yields the familiar sum and difference formulas

\[ \cos(x \pm y) = \cos{x} \cos{y} \mp \sin{x} \sin{y} \]

and

\[ \sin(x \pm y) = \sin{x} \cos{y} \pm \cos{x} \sin{y}. \]

An important corollary of Euler's theorem is de Moivre's theorem.

De Moivre's Theorem

\[ (\cos{x} + i \sin x)^\phi = \cos{\phi x} + i \sin{\phi x} \]

We have\[\begin{align*}(\cos{a\phi}+i\sin{a\phi}) \times (\cos{b\phi}+i\sin{b\phi}) &=\ \cos{a\phi}\cos{b\phi}+i\cos{a\phi}\sin{b\phi}+i\sin{a\phi}\cos{b\phi}-\sin{a\phi}\sin{b\phi} \\(\cos{a\phi}+i\sin{a\phi})(\cos{b\phi}+i\sin{b\phi}) &= \cos{a\phi}\cos{b\phi}-\sin{a\phi}\sin{b\phi}+i(\cos{a\phi}\sin{b\phi}+\sin{a\phi}\cos{b\phi}) \\\Rightarrow \cos{(a\phi+b\phi)}+i\sin{(a\phi+b\phi)} &= \cos{\big((a+b)\phi\big)}+i\sin{\big((a+b)\phi\big)}.\end{align*}\]For \(a = b\), we have \[(\cos{a\phi}+i\sin{a\phi})(\cos{a\phi}+i\sin{a\phi}) = \cos{(a\phi+a\phi)}+i\sin{(a\phi+a\phi)} = \cos{(2a\phi)}+i\sin{(2a\phi)}.\]This implies that \[\begin{align*}(\cos{a\phi}+i\sin{a\phi})^n &=\ \underbrace{(\cos{a\phi}+i\sin{a\phi}) \times \cdots \times (\cos{a\phi}+i\sin{a\phi})}_{n\text{ times}} \\&= \cos{\big(\underbrace{(a+\cdots +a)}_{n\ a\text{'s}}\phi\big)}+i\sin{\big(\underbrace{(a+\cdots +a)}_{n\ a\text{'s}}\phi\big)} \\\Rightarrow (\cos{a\phi}+i\sin{a\phi})^n &= \cos{(na\phi)}+i\sin{(na\phi)}.\end{align*}\]Thus, we have\[ (\cos{x} + i \sin x)^\phi = e^{ix\phi} = e^{i(\phi x)} = \cos{\phi x} + i \sin{\phi x}.\ _\square\]

De Moivre's theorem has many applications. As an example, one may wish to compute the roots of unity, or the complex solution set to the equation \( x^n = 1 \) for integer \( n \). Notice that \( e^{2\pi ki} \) is always equal to \( 1 \) for \( k \) an integer, so the \( n^\text{th} \) roots of unity must be

\[ e^{2\pi ki / n} = \cos\left(\frac{2\pi k}n\right) + i \sin\left(\frac{2 \pi k}n\right). \]

This process is akin to dividing the unit circle up into \( n \) equally spaced wedges.

Find the cube roots of unity.

The cube roots of unity are

• \( \cos\left(\frac{2 \pi }{3}\right) + i \sin\left(\frac{2 \pi }{3}\right) = -\frac{1}{2} + i \frac{\sqrt{3}}{2} \)
• \( \cos\left(\frac{4 \pi }{3}\right) + i \sin\left(\frac{4 \pi }{3}\right) = -\frac{1}{2} - i \frac{\sqrt{3}}{2} \)
• \( \cos\left(\frac{6 \pi }{3}\right) + i \sin\left(\frac{6 \pi }{3}\right) = 1.\ _\square \)