Introduction
In this post I want to show a few different ways of proving that $e^{i\theta} = \cos\theta + i\sin\theta$. It’s a cute illustration of how it’s often possible and rather cool to look at and solve problems in different ways.
Approach 1
The first technique is one I encountered toward the end of my tenure as a Calc 2 TA last semester as I was going over Taylor and MacLaurin series with my students. If we look at the MacLaurin series for $\sin \theta$ and $\cos\theta$ are
so
Now, let’s look at the MacLaurin series for $e^{i\theta}$. Because
we have
which, miraculously, is exactly what we get if we interleave (sum up) the terms of $\cos\theta + i\sin\theta$.
Approach 2
The next approach uses techniques from a first course in calculus. We first observe that if $e^{i\theta}$ is going to be the same as $\cos\theta + i\sin\theta$, then the fraction $\frac{cos\theta + i\sin\theta}{e^{i\theta}} = (\cos\theta + i\sin\theta)e^{-i\theta} = 1$. We will show that the second equality holds.
To do this, first define $f(\theta) = (\cos\theta + i\sin\theta)e^{-i\theta}$. Next, we take the (rather annoying) derivative of $f(\theta)$
If $f’(\theta) = 0$ at all values of $\theta$, $f$ is constant! Which constant? Let’s plug in $\theta = 0$ and find out!
If $f$ takes the value 1 when $\theta = 0$ and $f$ is constant, it must take the value 1 everywhere. To sum up, we have
Rearranging the second equality by cross multiplying, we see that Euler’s identity holds.
Approach 3
The last technique is my favorite. It uses a bit of linear algebra in concert with differential equations to produce what I think is the most illuminating proof of Euler’s identity. First, consider the differential equation
A differential equation is what is a kind of functional equation. Rather than trying to find the value of a real-valued variable, we are trying to find a function whose derivatives satisfy a given relationship. In this case, we want to find a function $f$ such that $f$’s second derivative is the same as $f$.
We first note that $\cos\theta$ and $\sin\theta$ are both solutions to this equation:
Because our differential equation involves second derivatives, it’s solutions are sort of analogous to those of a quadratic equation: which is to say, there are two! Formally, we say that the solution space is a vector space of dimension 2. If $\sin\theta$ and $\cos\theta$ are linearly independent solutions, they form a basis of the solution space, which means that every solution to our differential equation must can be written in the form the form $a\cos\theta + b\sin\theta$ for some constants $a,b$.
To see that $\sin\theta$ and $\cos\theta$ are indeed linearly independent, suppose that for all $\theta \in [0,2\pi]$, $a\cos\theta + b\sin\theta = 0$. Let’s pick a particular $\theta$ value in this interval. If $\theta = \pi/2$, then we have $a \cdot 0 + b \cdot 1 = b = 0$ (where the $=0$ at the end is because we supposed that $a\cos\theta + b\sin\theta = 0$ for all $\theta \in [0,2\pi]$). If we pick another, say $\theta = 0$, we have $a \cdot 1 + b \cdot 0 = a = 0.$
So far, we’ve shown that if $a\cos\theta + b\sin\theta = 0$, then we know that $a = b = 0$. This constitutes a proof that $\sin\theta$ and $\cos\theta$ are linearly independent. Because we said the solution space is of dimension 2, they are a basis for the solution space to our original equation.
We can separately observe that $e^{i\theta}$ is also a solution to our equation because
Because $\sin\theta$ and $\cos\theta$ form a basis, we know that
for some yet unknown constants $a,b$. Now it just remains to figure out what $a$ and $b$ are. (Can you see what they should be?)
To find $a$ and $b$, we note that if $f(\theta) = e^{i\theta}$, then $f(0) = 1$ and $f’(0) = i$. Using the first condition, we have
Using the second (taking the derivative of both sides before plugging 0 in), we have
Putting these both together, we have $e^{i\theta} = \cos\theta + i\sin\theta,$ which is what we wanted.