One thing we haven't drilled down upon yet is the similarities between the $\textrm{cis}\, \theta$ function (i.e., $\cos \theta + i \sin \theta$) and an exponential function. So that differences in notation don't get in the way of our seeing these, let us define an exponential function with base $b$ as
$$\exp_b(x) = b^x$$
Then, notice all of the following are true:$\newcommand{cis}{\textrm{cis}\,}$
$$\begin{array}{c|c}
\exp_b(\alpha + \beta) = \exp_b(\alpha) \cdot \exp_b(\beta) & \exp_b 0 = 1\\\hline
\cis(\alpha + \beta) = \cis(\alpha) \cdot \cis(\beta) & \cis 0 = 1
\end{array}$$
One might even wonder -- is there some base $b$ where $\cis(\theta) = \exp_b(\theta)$ for all $\theta$? At first blush, we would say this is impossible, as $\cis(\theta)$ mostly outputs complex values (except when $\theta$ is an integer multiple of $\pi$, while $\exp_b(\theta) = b^{\theta}$ outputs solely positive real values in our experience. However, the same could earlier have be said of $\cos \theta$ and $\sin \theta$ -- that is, until we saw in the solution of the cubic equation that we might need to consider cases where the cosine of some input was complex! (*...perhaps by using a complex input?*)

Let's run with this idea. Suppose we could generalize exponential functions in the same way, allowing them to work on a domain and co-domain of $\mathbb{C}$. Can we find a base $b$ that makes the $\exp_b$ and $\cis$ functions identical?

Strangely, the difference quotient applied to $\cis \theta$ can help here. Consider the following: $$\begin{array}{rcl} \displaystyle{\frac{\cis (\theta + h) - \cis (\theta)}{h}} &=& \displaystyle{\frac{\cis (\theta) \cdot \cis (h) - \cis (\theta)}{h}}\\ &=& \displaystyle{\cis (\theta) \cdot \frac{\cis (h) - 1}{h}}\\\\ &=& \displaystyle{\cis (\theta) \cdot \frac{(\cos h + i \sin h) - 1}{h}}\\\\ &=& \displaystyle{\cis (\theta) \cdot \left(\left(\frac{\cos h - 1}{h}\right) + \left(\frac{\sin h}{h}\right) \cdot i \right)} \end{array}$$ Now consider what happens to the difference quotient when $h$ gets very small. We can anticipate part of what we will get upon looking at the diagram below.

Notice that $\arg(\cis (\theta + h) - \cis(\theta))$, and thus the arg of the difference quotient (at least for small positive $h$) matches the direction the green arrow points. As the value of $h$ gets smaller and smaller, this green arrow gets arbitrarily close to pointing in the same direction as the blue ray does. As this direction is at a right angle to the blue radius and recalling that we can rotate something in the complex plane by $90^{\circ}$ counter-clockwise about the origin by multiplying by $i$, we expect to see that as $h$ gets very small, the difference quotient should approximate a multiple of $i \cdot \cis(\theta)$.

However, the magnitude the difference quotient approaches as $h$ becomes very small is not as clear.

Fortunately, we have another way to attack this. We can come to the same conclusion about the argument of the difference quotient *and* deduce its magnitude if we focus on what happens to the two sub-expressions found above, $(\frac{\cos h - 1}{h})$ and $(\frac{\sin h}{h})$, when $h$ gets small. The next two sections below address exactly this.

Consider the picture of the unit circle below.

Notice the large triangle (with all 3 colors) has base of length $1$ since the circle drawn is a unit circle, and -- upon noting this large triangle is similar to the reference triangle for $h$ -- has height $\tan h$. Thus, the large triangle has area $\frac{\tan h}{2}$.

The blue and yellow areas combined form a sector of the circle. Recalling the area of the entire unit circle is $\pi$, the blue and yellow areas must then sum to $\pi \cdot \frac{h}{2\pi} = \frac{h}{2}$

Finally, the blue area by itself corresponds to a triangle with base $1$ and height $\sin h$. As such the blue area is $\frac{\sin h}{2}$.

Putting these together leads to the string inequality $$\frac{\sin h}{2} \le \frac{h}{2} \le \frac{\tan h}{2}$$

Reciprocating all three expressions reverses the inequality symbols $$\frac{2}{\sin h} \ge \frac{2}{h} \ge \frac{2}{\tan h}$$

Then, multiplying all three expressions by $\frac{\sin h}{2}$ means (for positive $h$, anyways), that $$1 \ge \frac{\sin h}{h} \ge \cos h$$ meaning that $\frac{\sin h}{h}$ is always trapped between $1$ and $\cos h$. Had $h$ been negative, then we would have multiplied by a negative in the last step and the inquality symbols would have again reversed -- but $\frac{\sin h}{h}$ would have still ended up trapped between the same two values.

As $h$ gets very close to zero, $\cos h$ gets arbitrarily close to $1$. Consequently, we can make $\frac{\sin h}{h}$ as close as desired to $1$ by choosing $h$ small enough.

Importantly, if the generalized cosine function that works on a domain of complex numbers has the nice property of being *continuous everywhere* (a property first mentioned in the section on the Fundamental Theorem of Algebra), then the above is true even for complex values of $h$ whose magnitudes are very small. The essential ingredient for a function to be continuous everywhere is that you can keep outputs from that function arbitrarily close together by insisting the corresponding inputs are sufficiently close together. That said, one shouldn't worry too much about the details at this point -- all the intricacies of continuity will be studied in depth in calculus!

Now we turn our attention to the other subexpression seen in the difference quotient...

Note that with a well-chosen multiplication by $1$, we can introduce a factor of $(\sin h)$ into this value so that we might take advantage of our last result: $$\frac{\cos h - 1}{h} = \frac{(\cos h - 1)}{h} \cdot {\color{red}{\frac{(\cos h + 1)}{(\cos h + 1)}}} = \frac{\cos^2 - 1}{h(\cos h + 1)} = \frac{-\sin^2 h}{h(\cos h + 1)}$$ But then, we can pull this last expression apart into two factors -- the first of which can be made arbitrarily close to $1$ and the second staying arbitrarily close to $0$, provided $h$ is small enough. $$\frac{-\sin^2 h}{h(\cos h + 1)} = \left(\frac{\sin h}{h}\right) \cdot \left(\frac{-\sin h}{\cos h + 1}\right)$$ Consequently, for very small $h$: $$\frac{\cos h - 1}{h} \approx 1 \cdot 0 = 0$$

Similar to our earlier result involving the sine function, this approximation also holds when $h$ is complex but with very small magnitude -- again due to the notion of *continuity*, whose details we continue to omit lest we spoil too much in your calculus class! 😁

Picking up where we left off with the difference quotient of $\cis \theta$, we now see that the difference quotient for $\cis(\theta)$ can be made arbitrarily close to $i\cdot \cis \theta$ $$\begin{array}{rcl} \displaystyle{\frac{\cis (\theta + h) - \cis (\theta)}{h}} &=& \displaystyle{\cis (\theta) \cdot \left(\left(\frac{\cos h - 1}{h}\right) + \left(\frac{\sin h}{h}\right) \cdot i \right)}\\ &=& \displaystyle{\cis (\theta) \cdot \left(0 + 1 \cdot i \right)}\\ &=& i \cdot \cis (\theta) \end{array}$$ Our earlier argument was correct! We did discover the difference quotient of $\cis(\theta)$ for small values $h$ approximated a multiple of $i \cdot \cis(\theta)$

This is remarkable, but recall why we started looking at this in the first place. We wanted to know if there was some base $b$ where $\cis(\theta) = b^{\theta}$.

With this in mind, let us look for a value of $b$ whereby the difference quotient of $exp_b (\theta)$ behaves in the same way. That is to say, where for very small $h$, we have $$\frac{\exp_b(\theta + h) - \exp(\theta)}{h} \approx i \cdot b^{\theta}$$

But if the above were true, would we also expect all of the following to be true for sufficiently small $h$? $$\begin{array}{rcl} \displaystyle{\frac{b^{\theta + h} - b^{\theta}}{h}} &\approx& i \cdot b^{\theta}\\ b^{\theta + h} - b^{\theta} &\approx& i \cdot b^{\theta} \cdot h\\ b^{\theta} (b^h - 1) &\approx& i \cdot b^{\theta} \cdot h\\ b^h - 1 &\approx& i \cdot h\\ b^h &\approx& 1 + i \cdot h\\ b &\approx& (1 + i \cdot h)^{\frac{1}{h}}\\ \end{array}$$ Now, let $x=\displaystyle{\frac{1}{i \cdot h}}$, which implies both $\displaystyle{i \cdot h = \frac{1}{x}} \, \textrm{ and } \, \displaystyle{\frac{1}{h} = x \cdot i}$

Let us use these facts to rewrite the above approximation in $x$ and see where it leads: $$\begin{array}{rcl} b &\approx& \displaystyle{\left(1 + \frac{1}{x}\right)^{x \cdot i}}\\\\ &\approx& \displaystyle{\left(\left(1 + \frac{1}{x}\right)^x \right)^i} \end{array}$$

Recalling that all of the above must be true even when $h$ is complex, as long as its magnitude is very small, choose $h = \frac{-i}{n}$ for a sufficiently large integer $n$, to ensure that is the case. This means that $x = n$ and we can write an approximation for our mysterious base $b$ as
$$b \approx \left( \left( 1 + \frac{1}{n} \right)^n \right)^i$$
This last step allows us to approximate the (real) value being raised to the $i^{th}$ power above to whatever accuracy is desired by simply plugging in a sufficiently large value of $n$. Doing this for $n=10000000$ (one trillion), we discover that
$$b \approx 2.71828182\ldots$$
After this value's discovery by Swiss mathematician Jacob Bernoulli in 1683, the incredibly prolific mathematician Leonard Euler -- fellow countryman and family friend to the Bernoullis -- gave it a name. Calling it "$e$" in a letter he wrote to another mathematician, Christian Goldbach.^{†}

^{† : This does not relate to what we are currently discussing, but the reader might find it curious to know that Goldbach had conjectured in 1742 in one of his letters back to Euler that every even number greater than $2$ is the sum of two primes. Amazingly, this very famous claim remains unresolved even today.}

While it is a happy coincidence that $e$ is the first letter in the word "exponential", some suspect he named the value after himself. Most feel this would have been out of character. Eli Maor, in his book *$e$: the Story of a Number* remarks "[Euler] was an extremely modest man and often delayed publication of his own work so that a colleague or student of his would get due credit". Modesty aside, one could argue doing so would have been justified -- Euler was after all the first to prove $e$ is irrational by rewriting it as a convergent infinite sum of factorials:
$$e = \frac{1}{1} + \frac{1}{1} + \frac{1}{1 \cdot 2} + \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \cdots$$
But that's a story for another day. The truth may be even more prosaic: Euler was using the letter $a$ in some of his other mathematical work, and $e$ was the next vowel. Olivier Gerard in $1999$ posited a different explanation -- one made by his friend Etienne Delacroix de La Valette, who observed that $e$ could have been for the word *ein* (which means "one" in German) or possibly *Einheit* (which means "unity" in the same language). This better matches the sentence Euler uses to define it, where he notes it's logarithm is unity.

It was actually Euler, in his 1748 textbook"By the end of the 17th century we can say that much more than being a calculating device suitably well-tabulated, the logarithm function, very much on the model of the hyperbola-area, had been accepted into mathematics. When, in the 18th century, this geometric basis was discarded in favour of a fully analytical one, no extension or reformulation was necessary – the concept of "hyperbola-area" was transformed painlessly into "natural logarithm".

*You should catch a strong scent of "socks and shoes" in these last two conclusions made by Euler!*

Regardless of who did what when and who should get credit for realizing the importance of this strange value $e = 2.71828182\ldots$, that short name for this wonderfully "natural" base is good for us, as we can now write $\cis(\theta)$ even *more* briefly! (*Less ink used is always a good thing, right?*):
$$\cis(\theta) = e^{i\theta}$$

Interestingly, evaluating $\cis(\theta)$ when $\theta = \pi$ gives us one of more beautiful identities in all of mathematics -- one which relates $5$ of its most important values: $$e^{i\pi} + 1 = 0 \quad \textrm{(Euler's Identity)}$$ As an even more important consequence, we can now easily define $e^z$ for any complex number $z=a+bi$ and $a,b \in \mathbb{R}$: $$e^z = e^{a+bi} = e^a \cdot e^{bi} = e^a (\cos b + i \sin b)$$

One interesting thing about $e^{i\theta}$ (however you choose to denote it) is that you can write it as a sum of an even function, $\cos \theta$ and an odd one, $i \sin \theta$. $$e^{i\theta} = \cos \theta + i \sin \theta$$ Curiously, when a function can be expressed as a sum of an even function and an odd one, such a decomposition is unique! To see this, suppose a function $f$ can be expressed in the following way, where $E(x)$ is an even function, and $O(x)$ is an odd function. $$f(x) = E(x) + O(x)$$

Necessarily then, we have $f(-x) = E(-x) + O(-x)$. Taking advantage of the even and odd properties of $E(x)$ and $O(x)$ respectively, it must also be the case that $$f(-x) = E(x) - O(x)$$ Adding the expressions for $f(x)$ and $f(-x)$, and dividing by $2$ yields $$E(x) = \frac{f(x) + f(-x)}{2}$$ while subtracting them and dividing by $2$ tells us $$O(x) = \frac{f(x) - f(-x)}{2}$$ This unique decomposition of a function into the sum of even and odd components when such components exist brings a couple of immediate consequences:

First, noting that $e^{i\theta} = \cos \theta + i \sin \theta$ and recognizing $\cos \theta$ is an even function and $i \sin \theta$ is an odd function, we then know
$$\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2} \quad \textrm{ and } \quad \sin \theta = \frac{ie^{i\theta} - ie^{-i\theta}}{2i}$$
(*Note, we actually start with $i\sin \theta$ to derive the latter formula, but later divide both sides by $i$.*)