Euler's Formula on Complex Numbers

Euler's Formula

Maybe you remember learning about the infamous Leonhard Euler a while back....he's got another formula!

Euler's Formula states that $$e^{i\varphi}=\cos \varphi + i\sin \varphi$$ for any real number $\varphi.$

This formula is one of the most important contributions to complex analysis $-$ and it will be very helpful when you are trying to solve equations with complex numbers! Here's an illustration of how Euler's Formula works in the complex plane.

You already know how to plot sine and cosine on the unit circle when given an angle, $\varphi.$ Now, we can do the same thing, but in terms of $e!$ Given a real number, $\varphi,$ we can plot the complex number $e^{i\varphi}$ on the unit circle, since $e^{i\varphi}=\cos \varphi + i\sin \varphi.$ If you need a refresher on how to graph complex numbers, click here.

Since the complex number $e^{i\varphi}$ is on the unit circle, it has a magnitude of $1,$ and it makes an angle of $\varphi$ with the positive real axis. If you vary $\varphi,$ you can plot every point on the unit circle!

Basically, the big takeaway is that you can write $\cos \varphi + i\sin \varphi$ as $e^{i\varphi}.$ This will come in handy later, especially when we are learning how to put complex numbers in exponential form. For now, let's do a quick example to see if you feel comfortable with the formula.

To solve this problem, we simply need to plug $\varphi = \dfrac{\pi}{2}$ into Euler's Formula. Let's do it! $$\begin{align} e^{i\varphi}&=\cos \varphi + i\sin \varphi\\ \xrightarrow{\varphi = \frac{\pi}{2}} e^{\frac{\pi}{2}i}&=\cos \big(\frac{\pi}{2}\big) + i\sin \big(\frac{\pi}{2}\big)\\ &=0+i(1)\\ &=i \end{align}$$

So, using Euler's Formula, we have found that $e^{\frac{\pi}{2}i}=i.$

An interesting application of Euler's Formula is proving Euler's Identity, one of the most pivotal equations in mathematics, which states that $e^{i \pi}+1=0.$

$$\textbf{Proof of Euler's Identity}: e^{i \pi}+1=0$$

To prove Euler's Identity, we will use Euler's Formula: $$e^{i \theta} = \cos \theta + i\sin \theta$$

Let's substitute in $\pi$ for $\theta$ in Euler's Formula: $$\begin{align} e^{i \pi} &= \cos \pi + i\sin \pi\\ e^{i \pi} &= -1 + i(0)\\ e^{i \pi} &= -1\\ \Rightarrow e^{i \pi} +1&= 0 \end{align}$$

At this point, we understand expressions such as $a^b$, for real numbers $a$ and $b$. Now let's think about how we evaluate $a^z$, for a complex number $z$.

At first, this question may seem unmotivated and nonsensical, but extending exponents to the complex numbers has many uses in math and physics. It is a wonderful example of the power of math to unify different concepts. Complex exponents unify geometry and complex numbers, for example, and allow us to see a "bigger picture" when investigating AC circuits in physics.

Complex exponents do not make much sense given the traditional definition of exponents. However, as with other problems, we begin with what we do know. Any complex number $z$ can be written as $x+yi$, for real $x$ and $y$. As a result, $a^z$ becomes $a^{x+yi}$.

Now, we need a good definition for what this should be. We start from the basics: we still want $e^{p+q}=e^pe^q$ even for complex exponents. As a result,

\begin{equation} a^z=a^{x+yi}=a^x a^{yi}. \end{equation}

We can handle $a^x$, since $a$ and $x$ are real. But $a^{yi}$ just seems unreal.

Euler's equation (one of numerous) answers just this question. It is a jewel of mathematics. We claim that

\begin{equation} e^{i \theta}=\cos\theta+i\sin\theta. \end{equation}

At this point, you may already guess at the aforementioned connections between complex numbers and geometry. Let's look at where this equation comes from.

It is important to keep in mind that this is a definition for $e^{i \theta}$. It is, in fact, the best definition.

A knowledge of calculus helps to explain Euler's formula, but feel free to take this on faith for now (as this is Algebra 2). We have, from the theory of Taylor series, that \begin{equation} e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+... \end{equation} \begin{equation} \cos z=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}... \end{equation} \begin{equation} \sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}..., \end{equation}

for all complex numbers $z$.

Now, let $z=i \theta$ in the first equation above. Consider the sum of all the even powers of $\theta$ in this expansion.

The sum of the even powers of $\theta$ in $e^{i \theta}$ is exactly $\cos\theta$ ! Also verify that the sum of the odd powers of $\theta$ is $i\sin\theta$.

One of the wonderful things about this formula is that the trigonometric functions can be redefined in terms of complex exponents.

Using Euler's formula, we have that $e^{iz}=\cos z+ i \sin z$ and $e^{-iz}=\cos (-z)+i\sin(-z)=\cos z-i \sin z$. Adding and subtracting, respectively, shows that \begin{equation} \cos z = \frac{e^{iz}+e^{-iz}}{2} \end{equation} \begin{equation} \sin z = \frac{e^{iz}-e^{-iz}}{2}. \end{equation}

Since all the other basic trigonometric functions can be defined in terms of $\sin z$ and $\cos z$, we're done!

Now let's see how powerful these expressions are.

Application 1: Prove that $\cos(a+b)=\cos a \cos b- \sin a \sin b$.

Solution 1: Rewrite the RHS using the expressions for cosine and sine above. We have that \begin{equation} \cos a \cos b=\frac{(e^{ia}+e^{-ia})(e^{ib}+e^{-ib})}{4}= \frac{e^{i(a+b)}+e^{-i(a+b)}}{4}+\frac{e^{i(a-b)}+e^{-i(a-b)}}{4} \end{equation} \begin{equation} \sin a \sin b=\frac{(e^{ia}-e^{-ia})(e^{ib}-e^{-ib})}{4i^2}= -\frac{e^{i(a+b)}+e^{-i(a+b)}}{4}+\frac{e^{i(a-b)}+e^{-i(a-b)}}{4}. \end{equation}

Subtracting gives \begin{equation} \cos a \cos b-\sin a \sin b=\frac{e^{i(a+b)}+e^{-i(a+b)}}{2}=\cos(a+b), \end{equation}

as desired.

Similarly, we can show that $\sin(a+b)=\sin a \cos b + \sin b \cos a$. You are encouraged to work this out, as an exercise. You should also see the product to sum identities lurking in these proofs. Euler's identity offers an easy way to rederive these quickly.

Application 2: (De Moivre's Formula) Prove that $(\cos x+i \sin x)^n=\cos(nx)+i\sin(nx)$.

Proof: This is just another way to say $(e^{ix})^n=e^{i(nx)}$ !

De Moivre's Formula allows us to derive many other trigonometric multiple-angle identities. For instance, consider the expansion of $(e^{ix})^5=(\cos x + i \sin x)^5$:

\begin{equation} (e^{ix})^5=\cos^5 x+5i \cos^4 x \sin x -10 \cos^3 x \sin^2 x-10i \cos^2 x \sin^3 x+5 \cos x \sin^4 x +i\sin^5 x. \end{equation}

However, by De Moivre's Formula, we know that this is equivalent to $\cos 5x + i \sin 5x$. Thus, by equating real and imaginary parts, we obtain \begin{equation} \cos 5x=\cos^5 x-10 \cos^3 x+5 \cos x \sin^4 x \end{equation} \begin{equation} \sin 5x= 5 \cos^4 x-10 \cos^2 x \sin^3 x+ \sin^5 x. \end{equation}

Notice that the complex exponent encodes information for both sine and cosine identities together. We finish with a piece of notation: $\text{cis} x \cong \cos x + i \sin x$ (cis stands for Cosine plus I Sine).

Euler's Formula for Complex Numbers

We are familiar with Euler's Identity with rectangular coordinates, but something happens when we add complex numbers to the mix: the formula changes!

Remember that:

• $i$ signifies the imaginary number/imaginary component.

• We've incorporated the polar form of a complex number.

In the next lesson, you will learn about the exponential form of a complex number.

Euler's formula and continuous rotations

by All Angles

In this video by All Angles, you will learn more about Euler's Formula.