Homework Help: 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