# Showing that e^a * e^b = e^(a+b)

1. Jan 12, 2012

### dacruick

1. The problem statement, all variables and given/known data
By definition, $e^{z}$ = $\sum$$\frac{1}{n!}$$z^{n}$

Use this to show the relationship in the question title

3. The attempt at a solution
Well, what I've tried to do is as follows:
$e^{z_{1}}$$e^{z_{2}}$ = $\sum$$(z_{1}z_{2})^{n}$/${n!^{2}}$

And set that equal to
$e^{(z_{1}+z_{2})}$ = $\sum$$\frac{1}{n!}$$(z_{1}+z_{2})^{n}$

What I'm left with is this expression that
$(z_{1}+z_{2})^{n}$ = $(z_{1}z_{2})^{n}/n!$

those are all sums from n=0 to n=∞ still btw. Sorry for my LameTex inexperience.

My question is, is there any way that I can show that those two are equal? Or have I gone about this all wrong. Thanks!

dacruick

2. Jan 12, 2012

### Poopsilon

You can't multiply infinite series like you've done. In fact you can't even multiply finite polynomials like that. For example consider the multiplication of the following quadratics:

$$(x^2 + x + 1)(y^2 + y + 1) = (xy)^2 + x^2y + xy^2 + x^2 + y^2 + x + y + 1 \neq (xy)^2 + xy + 1$$

3. Jan 12, 2012

### dacruick

I'm so stupid...Haha, not just because of the foolishness you stated but because as soon as you said that I'm pretty sure I figured out how to do it. Maybe it had something to do with that cup of coffee I just made. You'll have to tussle with Starbucks for partial credit of my pending success.

4. Jan 12, 2012

### Poopsilon

Starbucks is far too successful already, I want all of it.

5. Jan 12, 2012