Euler's identity: Difference between revisions

Revision as of 10:23, 17 February 2016

Euler's identity is $e^{i\theta}=\cos \theta+ i\sin\theta$ . It is named after the 18th-century mathematician Leonhard Euler.

Background

Euler's formula is a fundamental tool used when solving problems involving complex numbers and/or trigonometry. Euler's formula replaces "cis", and is a superior notation, as it encapsulates several nice properties:

De Moivre's Theorem

De Moivre's Theorem states that for any real number $\theta$ and integer $n$ , $(\cos(\theta) + i\sin(\theta))^n = (e^{i\theta})^n = e^{in\theta} = \cos(n\theta) + i\sin(n\theta)$ .

Sine/Cosine Angle Addition Formulas

Start with $e^{i(\alpha + \beta)} = (e^{i\alpha})(e^{i\beta})$ , and apply Euler's forumla both sides:

$\cos(\alpha + \beta) + i \sin(\alpha + \beta) = (\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta).$

Expanding the right side gives

$(\cos\alpha\cos\beta - \sin\alpha\sin\beta) + i(\cos\alpha\sin\beta + \sin\alpha\cos\beta).$

Comparing the real and imaginary terms of these expressions gives the sine and cosine angle-addition formulas:

$\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$

$\sin(\alpha+\beta) = \cos\alpha\sin\beta + \sin\alpha\cos\beta$

Geometry on the complex plane

Other nice properties

A special, and quite fascinating, consequence of Euler's formula is the identity $e^{i\pi}+1=0$ , which relates five of the most fundamental numbers in all of mathematics: e, i, pi, 0, and 1.

Proof 1

The proof of Euler's formula can be shown using the technique from calculus known as Taylor series.

We have the following Taylor series:

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots=\sum_{k=0}^{\infty}\frac{x^k}{k!}$

$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots=\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k+1}}{(2k+1)!}$

$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots=\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k)!}$

The key step now is to let $x=i\theta$ and plug it into the series for $e^x$ . The result is Euler's formula above.

Proof 2

Define $z=\cos{\theta}+i\sin{\theta}$ . Then $\frac{dz}{d\theta}=-\sin{\theta}+i\cos{\theta}=iz$ , $\implies \frac{dz}{z}=id\theta$

$\int \frac{dz}{z}=\int id\theta$

$\ln{|z|}=i\theta+c$

$z=e^{i\theta+c}$ ; we know $z(0)=1$ , so we get $c=0$ , therefore $z=e^{i\theta}=\cos{\theta}+i\sin{\theta}$ .

@@ Line 6: / Line 6: @@
 ===De Moivre's Theorem===
-[[De Moivre's Theorem]] states that for any [[real number]]s <math>\theta</math> and <math>n</math>,
+[[De Moivre's Theorem]] states that for any [[real number]] <math>\theta</math> and integer <math>n</math>,
 <math>(\cos(\theta) + i\sin(\theta))^n = (e^{i\theta})^n = e^{in\theta} = \cos(n\theta) + i\sin(n\theta)</math>.

Page

Toolbox

Search