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Euler's formula

(Redirected from Eulers formula in complex analysis)

This article is about the Euler's formula in complex analysis. There is also Euler's formula which is related to the Euler characteristic in algebraic topology.

Euler's formula, named after Leonhard Euler, is a mathematical formula in the subfield of complex analysis that shows a deep relationship between the trigonometric functions and the complex exponential function. (Euler's identity is a special case of the Euler formula).

Euler's formula states that, for any real number x,

$e^{ix} = \cos x + i\;\sin x$

where

e is the base of the natural logarithm

i is the imaginary unit

sin and cos are trigonometric functions.

Contents

1 History

2 Notes

3 Proofs

3.1 Using Taylor series
3.2 Using calculus

4 External links

History

Euler's formula was proved (in an obscured form) for the first time by Roger Cotes in 1714, then rediscovered and popularized by Euler in 1748. It is interesting to note that neither of these men saw the geometrical interpretation of the formula: the view of complex numbers as points in the plane arose only some 50 years later (see Caspar Wessel).

Notes

This formula can be interpreted as saying that the function e^ix traces out the unit circle in the complex number plane as x ranges through the real numbers. Here, x is the angle that a line connecting the origin with a point on the unit circle makes with the positive real axis, measured counter clockwise and in radians. The formula is valid only if sin and cos take their arguments in radians rather than in degrees.

The proof is based on the Taylor series expansions of the exponential function e^z (where z is a complex number) and of sin x and cos x for real numbers x (see below). In fact, the same proof shows that Euler's formula is even valid for all complex numbers x.

The formula provides a powerful connection between analysis and trigonometry. It is used to represent complex numbers in polar coordinates and allows the definition of the logarithm for complex arguments. By using the exponential laws

$e^{a + b} = e^a \cdot e^{b}$

and

$(e^a)^b = e^{a b} \,$

(which are valid for all complex numbers a and b), one can also readily derive several trigonometric identities as well as de Moivre's formula from it. Euler's formula also allows one to interpret the sine and cosine functions as mere variations of the exponential function:

$\cos x = {e^{ix} + e^{-ix} \over 2}$

$\sin x = {e^{ix} - e^{-ix} \over 2i}$

These formulas can even serve as the definition of the trigonometric functions for complex arguments x. You can derive the two equations above simply by adding or subtracting Euler's formulas:

$e^{ix} = \cos x + i \sin x \;$

$e^{-ix} = \cos x - i \sin x \;$

and solving for either cosine or sine.

The formulae above can also be used to relate the hyperbolic sine and hyperbolic cosine functions to the usual trigonometric functions.

In differential equations, the function e^ix is often used to simplify derivations, even if the final answer is a real function involving sine and cosine. Euler's identity is an easy consequence of Euler's formula.

In electrical engineering and other fields, signals that vary periodically over time are often described as a combination of sine and cosine functions (see Fourier analysis), and these are more conveniently expressed as the real part of exponential functions with imaginary exponents, using Euler's formula.

Proofs

Using Taylor series

Here is a proof of Euler's formula using Taylor series expansions as well as basic facts about the powers of i:

$i^0=1 \,$

$i^1=i \,$

$i^2=-1 \,$

$i^3=-i \,$

$i^4=1 \,$

and so on. The functions e^x, cos(x) and sin(x) (assuming x is real) can be written as:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$

and for complex z we define each of these function by the above series, replacing x with iz. This is possible because the radius of convergence of each series is infinite. We then find that

$e^{iz} = 1 + iz + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \frac{(iz)^6}{6!} + \frac{(iz)^7}{7!} + \frac{(iz)^8}{8!} + \cdots$

$= 1 + iz - \frac{z^2}{2!} - \frac{iz^3}{3!} + \frac{z^4}{4!} + \frac{iz^5}{5!} - \frac{z^6}{6!} - \frac{iz^7}{7!} + \frac{z^8}{8!} + \cdots$

$= \left( 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \frac{z^8}{8!} + \cdots \right) + i\left( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots \right)$

$= \cos (z) + i\sin (z) \,$

The rearrangement of terms is justified because each series is absolutely convergent. Taking z = x to be a real number, gives the original identity as Euler discovered it.

Q.E.D.

Using calculus

Define the complex number $z$ such that

$z=\cos x + i\sin x \,$

Differentiating $z$ with respect to $x$ :

$\frac{dz}{dx}=-\sin x + i\cos x$

Using the fact that i² = -1:

$\frac{dz}{dx}=i^2\sin x + i\cos x=i(\cos x + i\sin x)=zi$

Separating variables and integrating both sides:

$\int\frac{1}{z}\,dz=\int i\,dx$

$\ln z=xi + C\,$

where

C

is the constant of integration.

To finish the proof we have to argue that it is zero. This is easily done by substituting $x = 0$ .

$\ln z = C\,$

But $z$ is just equal to:

$z = \cos x + i\sin x = \cos 0 + i \sin 0 = 1 \,$

thus

$\ln 1 = C \,$

$C = 0 \,$

So now we just exponentiate

$\ln z = xi \,$

$e^{\ln z} = e^{xi} \,$

$z = e^{xi} \,$

$e^{xi} = \cos x + i\sin x \,$

Q.E.D.

External links

Euler and his beautiful and extraordinary formula by Antonio Gutierrez from Geometry Step by Step from the Land of the Incas.
Euler's Formula - Puzzle: 55 pieces in a six star style of piece by Antonio Gutierrez from Geometry Step by Step from the Land of the Incas.

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10-26-2009 08:16:03
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