Euler’s formula

Proofs Euler’s formula using the MacLaurin series for sine and cosine. Introduces Euler’s identify and Cartesian and Polar coordinates.\(\)

Around 1740, the Swiss mathematician, physicist and engineer Leonhard Euler obtained the formula later named after him.

Euler’s Formula

Use the MacLaurin series for cosine and sine, which are known to converge for all real \(x\), and the MacLaurin series for \(\mathrm{e}^x\).

$$ \begin{align} \cos x &= \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots \\ \sin x &= \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots \\ \mathrm{e}^z &=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots \label{eq:ez} \end{align} $$

Substitute \(z=j\varphi\) in \(\eqref{eq:ez}\)

$$ \begin{align} \mathrm{e}^{j\varphi}&=\sum_{n=0}^{\infty}\frac{(j\varphi)^n}{n!}=1+j\varphi+\frac{(j\varphi)^2}{2!}+\frac{(j\varphi)^3}{3!}+\cdots \nonumber\\ &=1+j\varphi-\frac{\varphi^2}{2!}-j\frac{\varphi^3}{3!}+\frac{\varphi^4}{4!}+j\frac{\varphi^5}{5!}-\cdots \nonumber\\ &=\underbrace{\left(1-\frac{\varphi^2}{2!}+\frac{\varphi^4}{4!}+\cdots\right)}_{\cos\varphi} +j\underbrace{\left(\varphi-\frac{\varphi^3}{3!}+\frac{\varphi^5}{5!}-\cdots\right)}_{\sin\varphi} \end{align} $$

This leads us to Euler’s formula

$$ \shaded{ \mathrm{e}^{j\varphi}=\cos\varphi+j\sin\varphi } $$

Euler’s Identify

For the special case where \(\varphi=\pi\):

$$ \mathrm{e}^{j\pi} = \cos\pi+j\sin\pi = -1 $$

Rewritten as

$$ \shaded{ \mathrm{e}^{j\pi}+1 = 0 } $$

This combines many of the fundamental numbers with mathematical beauty

  • The number \(0\), the additive identify
  • The number \(1\), the multiplicative identity
  • The number \(\pi\), the ratio between a circle’s circumference and its diameter
  • The number \(j\), used to find the roots of polynomials defined as \(j=\sqrt{-1}\)
  • The number \(e\), from continuous compounding interest and from \(\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^x=\mathrm{e}^x\)

Cartesian and Polar coordinates

Euler’s formula traces out a unit circle in the complex plane as a function of \(\varphi\).  Here, \(\varphi\) is the angle that a line connecting the origin with a point on the unit circle makes with the positive real axis, measured in radians.

Relation between cartesian and polar coordinates (Euler)
Relation between cartesian and polar coordinates

A point in the complex plane can be represented by a complex number written in cartesian coordinates.  Euler’s formula lets you convert between cartesian and polar coordinates.  The polar form simplifies the mathematics when used in multiplication or powers of complex numbers. [wiki]

Any complex number \(z=x+jy\) can be written as

$$ \shaded{ z=x+jy=r(\cos\varphi+j\sin\varphi) = r\,\mathrm{e}^{j\varphi} } $$
where
$$ \begin{aligned} x&=\Re\{z\} & \mathrm{real\ part\ of\ }z\\ y&=\Im\{z\} & \mathrm{imaginary\ part\ of\ }z\\ r&=|z|=\sqrt{x^2+y^2} & \mathrm{magnitude\ of\ }z\\ \varphi&=\angle z=\mathrm{atan2}(y,x) & \mathrm{angle\ of\ }z \end{aligned} $$

Another notation method

$$ \shaded{r\,\mathrm{e}^{j\varphi} = \mathrm{e}^{\ln r}\mathrm{e}^{j\varphi} = \mathrm{e}^{\ln r+j\varphi}} $$

Relation to trigonometry

Euler’s formula connects analysis with trigonometry.  The connections most easy follow from adding or subtracting Euler’s formulas. [wiki]

$$ \left\{ \begin{aligned} \mathrm{e}^{j\varphi}&=\cos\varphi+j\sin\varphi \\ \mathrm{e}^{-j\varphi}&=\cos(-\varphi)+j\sin(-\varphi)=\cos\varphi-j\sin\varphi \end{aligned} \right. $$

Adding gives the \(\cos\varphi\)

$$ \left. \begin{aligned} \mathrm{e}^{j\varphi}&=\cos\varphi+j\sin\varphi \\ \mathrm{e}^{-j\varphi}&=\cos(-\varphi)+j\sin(-\varphi)=\cos\varphi-j\sin\varphi \end{aligned} \right\} \overset{add}{\Rightarrow} $$

Addition

$$ \begin{align} \mathrm{e}^{j\varphi}+\mathrm{e}^{-j\varphi} &= (\cos\varphi+\cancel{j\sin\varphi}) + (\cos\varphi-\cancel{j\sin\varphi}) \nonumber \\ &= 2\cos\varphi \end{align} $$

So

$$ \shaded{ \cos\varphi = \frac{\mathrm{e}^{j\varphi} + \mathrm{e}^{-j\varphi}}{2} } \label{eq:Eulers_sine} $$

Subtracting gives the \(\sin\varphi\)

$$ \left. \begin{align} \mathrm{e}^{j\varphi}&=\cos\varphi+j\sin\varphi \nonumber\\ \mathrm{e}^{-j\varphi}&=\cos(-\varphi)+j\sin(-\varphi)=\cos\varphi-j\sin\varphi\nonumber \end{align} \right\} \overset{subtract}{\Rightarrow} $$

Subtraction

$$ \begin{align} \mathrm{e}^{j\varphi}-\mathrm{e}^{-j\varphi} &= (\cancel{\cos\varphi}+j\sin\varphi) – (\cancel{\cos\varphi}-j\sin\varphi) \nonumber \\ &=2j\sin\varphi \end{align} $$

Thus

$$ \shaded{ \sin\varphi = \frac{\mathrm{e}^{j\varphi} – \mathrm{e}^{-j\varphi}}{2j} } \label{eq:Eulers_cosine} $$

This concludes “Euler’s formula and identify”.

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