Extension: Differential Proofs of Euler's Formula

Proof of Euler's Formula Using Differential Equations?

Let $y=\cos\theta+i\sin\theta$. Then $\frac{dy}{d\theta}=-\sin\theta+i\cos\theta=iy$. $y=e^{i\theta}$ also satisfies this. But this differential equation has only one unique solution, so $\cos\theta+i\sin\theta=e^{i\theta}$.

Tell me if my proof is wrong.

Recall the following identity, which makes the polar form of complex numbers very powerful: $$(r\mathrm{cis}\alpha)(s\mathrm{cis}\beta) = rs\mathrm{cis}(\alpha+\beta)$$ If we take a single number $z = r\mathrm{cis}\theta$ and raise it to a power $n$, we see that: $$(r\mathrm{cis}\theta)^n = r^n\mathrm{cis}(n\theta)$$ Setting $r=1$, we obtain a complex number of unit magnitude, $z = \cos\theta + i\sin\theta$. It follows that: $$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$$ This theorem is given its own name: De Moivre's Theorem. Note, however, that it does not need any special memorization, as it is a direct consequence of the general theorem for multiplication of any two complex numbers.

Notice that when you multiply two complex numbers, you add their arguments. Does this remind you of anything that you may already be familiar with? Think about this for a minute before continuing.

Think about exponentiation: when you multiply two numbers with the same base, you add their exponents! This may lead you to think that the argument of a complex number could act as an exponent of some form. But then ... what number should be our base?

Leonhard Euler proved that the argument of a complex number can indeed be used as an exponent. His theorem, known as Euler's Formula or Euler's Identity, stands as one of the most important theorems in mathematics. It states that: $$e^{i\theta} = \cos\theta + i\sin\theta$$ If your mind just got blown, don't worry; you are certainly not alone. However, all we need to prove this theorem is some calculus.

Consider the quantity $f(\theta)=e^{-i\theta}(\cos\theta+i\sin\theta)$. If we can prove that this quantity is always equal to $1$, then we will have successfully proven Euler's identity!

How do we prove that a function is constant? Think about this before continuing.

In this case, we can simply prove that the derivative of the function is always $0$! For this proof, we will assume that the process of taking a derivative is still legal for complex-valued functions, and that the process is exactly the same as that for taking a derivative for a real-valued function. It turns out that this assumption is indeed true, but proving it would require complex analysis.

Let us see what happens when we take the derivative of this function: $$f'(\theta) = -ie^{-i\theta}(\cos\theta+i\sin\theta) + e^{-i\theta}(-\sin\theta + i\cos\theta)$$ $$f'(\theta) = e^{-i\theta}(-i\cos\theta + \sin\theta - \sin\theta + i\cos\theta)$$ $$f'(\theta) = 0$$

Thus, the derivative of $f(\theta)$ is $0$, regardless of the value of $\theta$! Now, all we need to do is check the value of $f$ at a convenient value of $\theta$, and that value will be the value of $f$ everywhere!

Let's just pick $\theta=0$, because that seems the most convenient: $$f(0) = e^0(\cos 0 + i\sin 0)$$ $$f(0) = 1(1 + 0)$$ $$f(0) = 1$$

It follows that $f(\theta) = 1$ for all $\theta$. Therefore, $e^{i\theta} = \cos\theta + i\sin\theta$, and we are done!

A Short Proof of Euler's Identity

Euler's identity is regarded as the most beautiful equation in mathematics, since it combines five very important numbers $e,\pi,i,0,1$. If you are wondering why $0$ and $1$ are so important, here is a minor reason: $0$ is the identity of addition operation, $1$ is the multiplicative identity. That is, $a+(-a)=0$ and $a\cdot a^{-1}=1$. We will use only basic calculus and known trigonometry values.

The identity is: $$\begin{align*} e^{i\pi} +1 & = 0 \end{align*}$$ The identity is a special case of the following general one: $$\begin{align*} e^{ix} & = \cos x+i\sin x \end{align*}$$

It has many proofs. Here is an elegant one.

Let $y=\cos x+i\sin x$. Then $$\begin{align*} \dfrac{dy}{dx} & = -\sin x+i\cos x\newline & = i^2\sin x+i\cos x\newline & = i(\cos x+i\sin x)\newline & = iy\newline \dfrac{dy}y & = idx \end{align*}$$ Now let's integrate both sides. As we know $\int \frac{1}x=\ln x$ and $\int dx = x$, we have $$\begin{align*} \int \dfrac{dy}y & = \int idx\newline \ln y & = ix+C \end{align*}$$ for some constant $C$ (since it is indefinite integration). We get $$\begin{align*} e^{ix+C} & = y\newline & = \cos x+i\sin x \end{align*}$$ We need to find the value of $C$, so let's set $x=0$, and we get $$\begin{align*} e^{C} & = 1 \end{align*}$$ So $C=0$ and we have $$\begin{align*} e^{ix} & = \cos x+i\sin x \end{align*}$$ Set $x=\pi$ and we have $$\begin{align*} e^{i\pi} +1 & = 0 \end{align*}$$ DONE!

A Quick Way to Verify that Euler's Formula is Correct

We're going to start our post by stating Euler's Formula, which says that

$$e^{i\theta}=\cos\theta+i\sin\theta$$

How do we know that this formula is correct? Here's a quick way to do that.

Define a function $f(\theta)=e^{-i\theta}(\cos\theta+i\sin\theta)$.

We're going to take the derivative of this function with respect to $\theta$.

But in order to that, we need to know the product rule which says that

$$(f(x) \cdot g(x))'=f(x)\cdot g(x)'+f(x)'\cdot g(x)$$

So,

$$f'(\theta)=e^{-i\theta}\cdot(-\sin\theta+i\cos\theta)+(-ie^{-i\theta})\cdot(\cos\theta+i\sin\theta)$$ $$\quad\quad\quad = e^{-i\theta}\cdot(-\sin\theta+i\cos\theta)-e^{-i\theta}\cdot(-\sin\theta+i\cos\theta)=0$$

Also notice that $f(0)$ is equal to $1$, since

$$e^{-i(0)}(\cos(0) + i\sin(0)) = 1(1 + 0i) = 1.$$

So, this means that $f(\theta)$ is always equal to $1$.

In other words,

$$\frac{\cos\theta+i\sin\theta}{e^{i\theta}}=1$$

Or, $$e^{i\theta}=\cos\theta+i\sin\theta$$

which proves Euler's formula.

Note: Technically, we first need to show that the rules of differentiation hold over the complex numbers. However, that requires knowledge of complex analysis and, for now, we can make the assumption that it is true.

Euler's Formulas states that

Let $\theta$ be an angle in radians. Then $$e^{i\theta} = \cos \theta + i\sin \theta$$

If you're seeing this the first time in your life, then you'll be wondering "how in the world do we have a complex exponent?" Here's how

Now factor out that $i$. Notice anything interesting?

We get $f'(\theta) = i(\cos \theta + i\sin\theta) = f(\theta)$. Thus, we conclude that $i = \frac{f'(\theta)}{f(\theta)}$. Let's integrate both sides with respect to $\theta$.

Then $$i\theta = \log f(\theta) + C$$ We easily find that $C = 0$ and $f(\theta) = e^{i\theta} = \cos \theta + i \sin \theta$. $\Box$