# Introduction

Euler's Formula

Euler's Formula, sometimes called Euler's identity, states that[1]:

$e^{ix}=\cos(x)+i\sin(x).$

Read more about the imaginary number i in the helper page Complex Numbers. This result was discovered by Leonhard Euler around 1740[1]. After replacing variable x with constant π, the formula turns into:

$e^{i\pi}+1=0$

which is considered one of the most beautiful mathematical formulas of all time. It connects five constants of mathematics and also the three most important mathematical operations: addition, multiplication, and exponentiation.

To Benjamin Peirce, one of Harvard's leading mathematicians in the nineteenth century, this formula came as something of a revelation. Having discovered it one day, he turned to his students and said[2]: "Gentlemen, that is surely true, it is absolutely paradoxical; we cannot understand it, and we don't know what it means. But we have proved it, and therefore we know it must be the truth."

# A Geometric Representation of the Right Hand Side

Figure 1: Click to enlarge

In this article we take a geometric approach to the formula. Recall that the parametric representation in terms of the sine and cosine functions of a unit circle is:

$\sin^2x+ \cos^2x=1$,

with a random point A being (cosθ, sinθ) (as shown in Figure 1), where θ is the angle formed by the positive x axis and the line connecting the origin O and point A We first try to form a visualization of right hand side, (cosx + isinx) on some coordinate system.

Figure 2: The complex plane

Yet after i is included, there's no plausible representation of (cosθ,isinθ) on R2. [3]By the end of the 18th century several mathematicians in rapid succession found the geometric representation of the then mystical quantity of a + bi as a point on a plane, or a vector connecting the origin to that point. The plane is denoted as C, as shown in Figure 2. The horizontal axis is denoted the real axis while the vertical axis is the imaginary axis.

With this in mind, we could replace a with cosx and b with sinx and equate the quantity to 1, and once again, we arrive at a parametrization of the unit circle, only this time on the complex plane C.

The right hand side of eix =cosx + isinx simply means a point on the unit circle in the complex plane with angle x, and if we take Euler's Formula for granted, we arrive at the conclusion that the left hand side, eix represents the same point. But why? What exactly does eix mean?

# Moving Particle Argument[4]

Figure 3: Click to Enlarge

To make sense of the left hand side of Euler's formula, eix, we first attempt a moving particle argument. We define eit to be position function of some particle with respect to time,

$Z(t)=e^{it}$

Recall the basic fact that $e^{x}$ is its own derivative:

$\frac{d}{dt}e^{x}=e^{x}$

Let's now "extend" the action of derivation from with just real numbers to include complex numbers, so that we can "assume":

$\frac{d}{dt}e^{it}=ie^{it}$

Note: An important generalization of the derivative concerns complex functions of complex variables, such as functions from (a domain in) the complex numbers C to C. The notion of the derivative of such a function is obtained by replacing real variables with complex variables in the definition. If C is identified with R2 by writing a complex number z as x + i y, then a differentiable function from C to C is certainly differentiable as a function from R2 to R2 (in the sense that its partial derivatives all exist), but the converse is not true in general.[5] For the purpose of this page, we will simply treat the complex constants like how constants would have been treated in real number derivations.

We are used to visually thinking of the derivative of a real function as the slope of the tangent to the graph of the function, but how are we to make sense of the derivation in the above function?

We appeal to a geometric interpretation. As in Figure 3, at time t the particle's position is given by Z(t). Next recall from physics that the velocity V(t)[6] is the vector whose length and direction are given by the instantaneous speed, and the instantaneous direction of motion (tangent to the trajectory), of the moving particle. The figure shows the movement M of the particle between time t and t + a, and this should make it clear that:

$\frac {d}{dt} Z(t)= \lim_{a \to 0}\frac{Z(t+a)-Z(t)}{a}=\lim_{a \to 0}\frac{M}{a}=V(t).$

Thus, given a complex function Z(t) of a real variable t, we can always visualize Z(t) as the position of a moving particle and its first derivative as its velocity. In our case of Z(t)=eit,

$V(t)=iZ(t)$

Figure 4: Movement of Particle

Let's take a second here to notice the relationships between the location vector Z(t) and the velocity vector V(t). We first take their dot product,

$2 Z(t)\cdot V(t)= Z(t)\frac{dZ(t)}{dt}+ Z(t)\frac{dZ(t)}{dt}$

Yet we notice the right hand side of the previous equation is:

$Z(t)\frac{dZ(t)}{dt}+ Z(t)\frac{dZ(t)}{dt}=\frac{d}{dt}Z(t)\cdot Z(t)$

So the dot product of Z(t) and V(t) is actually:

$Z(t)\cdot V(t)=\frac{1}{2} \frac{d}{dt}|Z(t)|^2$

Since Z(t) is a vector on a circle, its magnitude is a constant. Thus,

$Z(t)\cdot V(t)=\frac{1}{2} \frac{d}{dt}Constant=0$

Next, let's compare their magnitudes,

$|Z(t)|=e^{it}$

$|V(t)|=\sqrt{(ie^{it})^2}=e^{it}$

Thus,we have arrived at the following two conclusions:

• The magnitude of the position vector and velocity vector of the moving particle are equal.
• The position vector and the velocity vector are always perpendicular to each other.

Recall that in the Cartesian coordinate system a unit circle is a circle with the origin as its center and radius 1. According to the rules of complex transformation[7], to multiply a vector on the complex plane by i is to keep its length constant, while rotating it counterclockwise 90 degrees. Since the particle's initial position, Z(0), is equal to 1, its initial velocity is i, so it is moving vertically upward. A split second later, the particle will have moved very slightly in this direction, and its new velocity will be at right angles to its new position vector. Continuing to construct the motion in this way, it is clear that the particle will travel around the unit circle. This whole process it illustrated in Figure 4.

Since we now know that the length of the vector |Z(t)| remains equal to 1 throughout the motion, it follows the particle's speed (defined to be the length of the velocity vector) V(t) also remains equal to 1. Thus after time t=θ, the particle will have traveled a distance θ around the unit circle. This argument gives a geometric statement of the left hand side of the Euler's formula.

# Power Series Argument

As an alternative approach, we look at the left hand side of Euler's formula using power series.

Recall that the Taylor Series of a real function f(x) that is infinitely differentiable in a neighborhood of a real or complex number a is the power series[7]

$f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...$

Note: Since the Taylor series is only defined for real functions, our approach which concerns the imaginary unit as a constant will not be a rigorous proof of Euler's formula. However, we shall assume just as we did with the moving particles argument that i can be treated as merely a constant.

Since the defining property of the exponential function f(x)=ex is,

$\frac{d}{dt}e^x=e^x$

By setting a=0, we have:

$f'(a)=f''(a)=f'''(a)=...=e^{a}=1$

By substitution, the Taylor series of the left hand side of the Euler's formula becomes:

$f(x)=e^{ix}=f(a)+\frac{f'(a)}{1!}(ix-0)+\frac{f''(a)}{2!}(ix-0)^2+\frac{f'''(a)}{3!}(ix)^3+...$

Equa (1)         $f(x)=e^{ix}=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+...$

Recall some basic facts about the powers of i:

$i^{4n+1}=i \qquad i^{4n+2}=-1 \qquad i^{4n+3}=-i \qquad i^{4n}=1$

By substituting these four properties into Equa (1),

$e^{ix}=C(x)+iS(x)$, where

$C(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-... \qquad S(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$

By applying Taylor's Theorem we can also obtain the power series of the sine and cosine function[7],

$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-... \qquad \cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-...$

Thus, the left hand side of Euler's formula turns into:

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

# Applications of Euler's Formula[4]

## Sine and Cosine in terms of Euler's Formula

Figure 4: Sine and Cosine Defined Alternatively

A simple but important consequence of Euler's formula is that sine and cosine can be "redefined" in terms of the exponential function. We can visualize sine and cosine on the unit circle.

There's also a direct mathematical derivation for this:

Equa (2)         $e^{ix}=\cos x+i\sin(x)$
Equa (3)         $e^{-ix}=\cos(x)-i\sin(x)$

Then Equa(2) + Equa(3) and Equa(2) - Equa(3) yields:

$\cos x=\frac{e^{ix}+e^{-ix}}{2},\ \quad \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$

## Further Applications in Trigonometry

If in high school you were constantly baffled by the trigonometric identities that were simply impossible to remember, Euler's formula provides a new method to view them via complex multiplication. In other words, all trigonometric identities can be derived from complex multiplication. In the following examples, I will reduce clutter by using the following simplifications:

Figure 5: e^(i(theta+phi) on the unit circle)
 $C$ $=$ $\cos(\theta)$ $S$ $=$ $\sin(\theta)$ $c$ $=$ $\cos(\phi)$ $s$ $=$ $\sin(\phi)$

### Cos(θ+φ) and Sin(θ+φ)

We shall start by tackling the identity for cos(θ+φ), which you might have known the answer to. (Cc-Ss) With Euler's Formula, we easily relate cos(θ+φ) to real part of the point B, ei(θ+φ) on the unit circle, as shown in Figure 5.

$e^{i(\phi+\theta)}= e^{i\theta}e^{i\phi}$

$= (C+iS)(c+is)$
$= (Cc-Ss)+i(Sc+Cs)$

It seems we obtain not only the identity for cos(θ+φ), but also one for sin(θ+φ):

• $\cos(\theta+\phi)=\cos(\theta)cos(\phi)-\sin(\theta)\sin(\phi)$
• $\sin(\theta+\phi)=\sin(\theta)\cos(\phi)+\cos(\theta)+\sin(\phi)$

It's like two birds with one stone, isn' it?

### From Cos(3θ),Sin(3θ) to Cos(nθ),Sin(3θ)

With the previous derivation, it is natural to relate cos3θ and sin3θ to the real and imaginary components of ei3θ on the complex plane. The mathematical manipulation is as follows:

$e^{i3\theta}=(e^{i\theta})^3$

$=(C+iS)^3$
$=(C^3-3CS^2)+i(3C^2S-S^3)$

Recall the fundamental trigonometry identity,

$\sin^2\theta+\cos^2\theta=1$

Thus,

$(C^3-3CS^2)+i(3SC^2-S^3)=(4C^3-3C)+i(-4S^3+3S)$

We have derived our two identities,

• $\sin(3\theta)=3C^3-3C$
• $\cos(3\theta)=-4S^3+3S$

Now if we replace the constant integer 3 with a variable n, we will be able to generalize our results,

$e^{in\theta}=\cos(n\theta)+i\sin(n\theta)$

$=(\cos(\theta)+\sin(\theta))^n$

For low n values, simply plug in and expand the above general formula to get the corresponding trigonometry identity.

### Cos4θ

Using Euler's Formula, we can not only express trig functions of multiples of θ in terms of powers of trig functions of θ, but we can also go in the opposite direction. For example,

Since

$2\cos(\theta)=e^{i\theta}+e^{-i\theta}$

Then we take both sides to the fourth power to get:

$2^4(\cos(\theta))^4=(e^{i\theta}+e^{-i\theta})^4$

$=(e^{i4\theta}+e^{-i4\theta})+4(i^{2i\theta}+e^{-i2\theta})+6$
$=2\cos(4\theta)+8\cos(2\theta)+6$

# References

1. 1.0 1.1 Wikipedia. http://en.wikipedia.org/wiki/Euler's_formula
2. Science News. http://www.sciencenews.org/view/generic/id/8406/title/Math_Trek__Eulers_Beautiful_Equation
3. Wikipedia. This information is compiled from the Wiki link of the Complex Plane, http://en.wikipedia.org/wiki/Complex_plane.
4. 4.0 4.1 Author, Needham, Tristan. (1999). Complex Visual Analysis. Place of publication: Oxford University Press, USA. ISBN-13: 978-0198534464 The idea and main structure for this proof came from this book. I rewrote some of its proofs.
5. Wikipedia, http://en.wikipedia.org/wiki/Derivative
6. Wikipedia, http://en.wikipedia.org/wiki/Velocity
7. 7.0 7.1 7.2 This site has a clear, comprehensive overview of simple complex transformation rules, http://www.clarku.edu/~djoyce/complex/plane.html. This content could also be found in typical complex analysis texts.