Difference between revisions of "Euler's identity"

Latest revision as of 10:19, 15 February 2025

Euler's Formula 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 $\frac{dz}{d\theta}=-\sin{\theta}+i\cos{\theta}=iz$ , $\implies \frac{dz}{z}=id\theta$ $\int \frac{dz}{z}=\int id\theta$ $\ln{|z|}=i\theta+c$ $z=e^{i\theta+c}<math>; we know </math>z(0)=1<math>, so we get </math>c=0<math>, therefore </math>z=e^{i\theta}=\cos{\theta}+i\sin{\theta}$ .

@@ Line 9: / Line 9: @@
 <math>(\cos(\theta) + i\sin(\theta))^n = (e^{i\theta})^n = e^{in\theta} = \cos(n\theta) + i\sin(n\theta)</math>.
-===Sine/Cosine Angle Addition Formulas===
+=== Sine/Cosine Angle Addition Formulas ===
 Start with <math>e^{i(\alpha + \beta)} = (e^{i\alpha})(e^{i\beta})</math>, and apply Euler's forumla both sides:
+<cmath>\cos(\alpha + \beta) + i \sin(\alpha + \beta) = (\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta).</cmath>
-<math>
-\cos(\alpha + \beta) + i \sin(\alpha + \beta) = (\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta).</math>
 Expanding the right side gives
+<cmath>(\cos\alpha\cos\beta - \sin\alpha\sin\beta) + i(\cos\alpha\sin\beta + \sin\alpha\cos\beta).</cmath>
-<math>
-(\cos\alpha\cos\beta - \sin\alpha\sin\beta) + i(\cos\alpha\sin\beta + \sin\alpha\cos\beta).
-</math>
 Comparing the real and imaginary terms of these expressions gives the sine and cosine angle-addition formulas:
+<cmath>\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta</cmath>
-<math>
+<cmath>\sin(\alpha+\beta) = \cos\alpha\sin\beta + \sin\alpha\cos\beta</cmath>
-\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta
-</math>
-<math>
-\sin(\alpha+\beta) = \cos\alpha\sin\beta + \sin\alpha\cos\beta
-</math>
 ===Geometry on the complex plane===
@@ Line 36: / Line 23: @@
 ===Other nice properties===
-A special, and quite fascinating, consequence of Euler's formula is the identity <math>e^{i\pi}+1=0</math>, which relates five of the most fundamental numbers in all of mathematics: [[e]], [[imaginary unit | i]], [[pi]], [[zero (constant)| 0]], and 1.
+A special, and quite fascinating, consequence of Euler's formula is the identity <math>e^{i\pi}+1=0</math>, which relates five of the most fundamental numbers in all of mathematics: [[e]], [[imaginary unit|i]], [[pi]], [[0]], and [[1]].
 ==Proof 1==
@@ Line 43: / Line 30: @@
 We have the following Taylor series:
+<cmath>e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots=\sum_{k=0}^{\infty}\frac{x^k}{k!}</cmath>
-<math>e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots=\sum_{k=0}^{\infty}\frac{x^k}{k!}</math>
+<cmath>\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots=\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k+1}}{(2k+1)!}</cmath>
+<cmath>\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots=\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k)!}</cmath>
-<math>\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots=\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k+1}}{(2k+1)!}</math>
-<math>\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots=\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k)!}</math>
 The key step now is to let <math>x=i\theta</math> and plug it into the series for <math>e^x</math>.  The result is Euler's formula above.
@@ Line 54: / Line 38: @@
 ==Proof 2==
 Define <math>z=\cos{\theta}+i\sin{\theta}</math>. Then <math>\frac{dz}{d\theta}=-\sin{\theta}+i\cos{\theta}=iz</math>, <math>\implies \frac{dz}{z}=id\theta</math>
+<cmath>\int \frac{dz}{z}=\int id\theta</cmath>
+<cmath>\ln{|z|}=i\theta+c</cmath>
+<cmath>z=e^{i\theta+c}<math>; we know </math>z(0)=1<math>, so we get </math>c=0<math>, therefore </math>z=e^{i\theta}=\cos{\theta}+i\sin{\theta}</cmath>.
-<math>\int \frac{dz}{z}=\int id\theta</math>
+== See Also ==
-<math>\ln{|z|}=i\theta+c</math>
-<math>z=e^{i\theta+c}</math>; we know <math>z(0)=1</math>, so we get <math>c=0</math>, therefore <math>z=e^{i\theta}=\cos{\theta}+i\sin{\theta}</math>.
-== See Also ==
 *[[Complex numbers]]
 *[[Trigonometry]]
@@ Line 68: / Line 50: @@
 [[Category:Complex numbers]]
+{{stub}}

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