Time to show you the dramatic conclusion to the story I began yesterday.

Our problem was to define the complex exponential function in a way that was consistent with everything we knew about real exponential functions. We noticed that one of the standard rules for exponents implies

\[

e^{x+iy}=e^xe^{iy}.

\]

Since we already know how to deal with the first term in that product on the right-hand side, our problem has been reduced to deciding what to do with “pure imaginary” exponents.

The key to doing that is to remember that we have a Taylor series expansion for the exponential function:

\[

e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\dots,

\]

which is a pretty result all by itself. We have previously discussed Taylor series, so if you need a refresher course just follow the link.

Now, the cool thing about that series is that it is just as meaningful if we use a complex number in place of *x*. (That’s not so obvious, actually, and a rigorous course in complex variables would spend some time establishing that complex power series really are acceptable, but we will obscure those details.) We can write:

\[

e^{iy}=1+iy+\frac{(iy)^2}{2!}+\frac{(iy)^3}{3!}+\frac{(iy)^4}{4!}+\frac{(iy)^5}{5!}+\dots

\]

Let us now pause to recall some of the basic algebra of complex numbers:

\[

i^1=i, \phantom{xx} i^2=-1, \phantom{xx} i^3=i^2(i)=-i, \phantom{xx} i^4=i^2(i^2)=1.

\]

That’s very interesting, since it implies that raising *i* to any even power produces an integer. We will put that to good use by separating out the terms with even exponents from the series above:

\[

1+\frac{(iy)^2}{2!}+\frac{(iy)^4}{4!}+\frac{(iy)^6}{6!}+\frac{(iy)^8}{8!}+\dots

\]

Substituting in the even powers of *i* we saw above gives us:

\[

1-\frac{y^2}{2!}+\frac{y^4}{4!}-\frac{y^6}{6!}+\frac{y^8}{8!}+\dots

\]

If you remember your second semester calculus, that probably looks familiar. It is precisely the Taylor series for *cos y*.

Now let’s try the same trick for the terms with odd number exponents. We have:

\[

iy+\frac{(iy)^3}{3!}+\frac{(iy)^5}{5!}+\frac{(iy)^7}{7!}+\frac{(iy)^9}{9!}+\dots

\]

Applying our algebraic rules and then factoring out the *i* from every term leads to

\[

i \left(y-\frac{y^3}{3!}+\frac{y^5}{5!}-\frac{y^7}{7!}+\frac{y^9}{9!}-\dots \right)

\]

And if *that* looks familiar, it is because the expression in parentheses is precisely the Taylor series for *sin y*.

*That’s* how we arrive at the definition

\[

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

\]

which leads immediately to Euler’s identity

\[

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

\]

Very nice, I’d say, and very unexpected. Of course, it gets even better when you realize that Euler’s identity is really just a special case of a more general result, that the sum of all the *n*th roots of unity is zero for any *n*. But we’ll save that for a later post.