# Calculators and complex numbers (Part 17)

In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.

Definition. If $z$ is a complex number, then we define

$e^z = \displaystyle \sum_{n=0}^{\infty} \frac{z^n}{n!}$

Even though this isn’t the usual way of defining the exponential function for real numbers, the good news is that one Law of Exponents remains true. (At we saw in an earlier post in this series, we can’t always assume that the usual Laws of Exponents will remain true when we permit the use of complex numbers.)

Theorem. If $z$ and $w$ are complex numbers, then $e^z e^w = e^{z+w}$.

In yesterday’s post, I gave the idea behind the proof… group terms where the sums of the exponents of $z$ and $w$ are the same. Today, I will formally prove the theorem.

The proof of the theorem relies on a principle that doesn’t seem to be taught very often anymore… rearranging the terms of a double sum. In this case, the double sum is

$e^z e^w = \displaystyle \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{z^n}{n!} \frac{w^k}{k!}$

This can be visualized in the picture below, where the $x-$axis represents the values of $k$ and the $y-$axis represents the values of $n$. Each red dot symbolizes a term in the above double sum. For a fixed value of $n$, the values of $k$ vary from $0$ to $\infty$. In other words, we start with $n =0$ and add all the terms on the line $n = 0$ (i.e., the $x-$axis in the picture). Then we go up to $n = 1$ and then add all the terms on the next horizontal line. And so on.

I will rearrange the terms as follows: Let $j = n+k$. Then for a fixed value of $j$, the values of $k$ will vary from $0$ to $j$. This is perhaps best described in the picture below. The value of $j$, the sum of the coordinates, is constant along the diagonal lines below. The value of $k$ then changes while moving along a diagonal line.

Even though this is a different way of adding the terms, we clearly see that all of the red circles will be hit regardless of which technique is used for adding the terms.

In this way, the double sum $\displaystyle \sum_{n=0}^\infty \sum_{k=0}^\infty$ gets replaced by $\displaystyle \sum_{j=0}^\infty \sum_{k=0}^j$. Since $n = j-k$, we have

$e^z e^w = \displaystyle \sum_{j=0}^{\infty} \sum_{k=0}^{j} \frac{z^{j-k}}{(j-k)!} \frac{w^k}{k!}$

We now add a couple of $j!$ terms to this expression for reasons that will become clear shortly:

$e^z e^w = \displaystyle \sum_{j=0}^{\infty} \sum_{k=0}^{j} \frac{j!}{j!} \frac{1}{k! (j-k)!} w^k z^{j-k}$

Since $j!$ does not contain any $k$s, it can be pulled outside of the inner sum on $k$. We do this for the $j!$ in the denominator:

$= \displaystyle \sum_{j=0}^\infty \frac{1}{j!} \sum_{k=0}^j \frac{j!}{k!(j-k)!} w^k z^{j-k}$

We recognize that $\displaystyle \frac{j!}{k! (j-k)!}$ is a binomial coefficent:

$= \displaystyle \sum_{j=0}^\infty \frac{1}{j!} \sum_{k=0}^j {j \choose k} w^k z^{j-k}$

The inner sum is recognized as the formula for a binomial expansion:

$= \displaystyle \sum_{j=0}^\infty \frac{1}{j!} (w+z)^j$

Finally, we recognize this as the definition of $e^{w+z}$, using the dummy variable $j$ instead of $n$. This proves that $e^z e^w = e^{z+w}$ even if $z$ and $w$ are complex.

Without a doubt, this theorem was a lot of work. The good news is that, with this result, it will no longer be necessary to explicitly use the summation definition of $e^z$ to actually compute $e^z$, as we’ll see tomorrow.

For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.

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