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Euler's identity

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

Contents

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 numbers \theta and 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}.

See Also

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