# [SOLVED] Complex exponential proof

• Sep 15th 2008, 03:40 PM
gidget
[SOLVED] Complex exponential proof
I need the proof for (e^z1)(e^z2)=e^(z1+z2)
using the power series expansion for e^z=1+z+(z^2)/2! + (z^3)/3! +....
I've tried it a couple of times but cant seem to get it.
thanks for any help!!
• Sep 15th 2008, 03:46 PM
icemanfan
$\displaystyle e^{z1}e^{z2} = e^{z1+z2}$ is a result of one of the rules of exponents. It requires no proof, at least not if I'm teaching.
• Sep 15th 2008, 03:49 PM
gidget
yeah i've proved it using exponent laws no problem but our prof said to do it the other way and wants it proved using the power series.
• Sep 15th 2008, 04:49 PM
ThePerfectHacker
Quote:

Originally Posted by icemanfan
$\displaystyle e^{z1}e^{z2} = e^{z1+z2}$ is a result of one of the rules of exponents. It requires no proof, at least not if I'm teaching.

Not really. The exponent rule $\displaystyle a^ba^c=a^{b+c}$ works for $\displaystyle a,b,c>0$. But you cannot use this rule here because $\displaystyle z_1,z_2$ are complex numbers. Thus, would does it mean to raise an exponent?

Quote:

Originally Posted by gidget
I need the proof for (e^z1)(e^z2)=e^(z1+z2)
using the power series expansion for e^z=1+z+(z^2)/2! + (z^3)/3! +....
I've tried it a couple of times but cant seem to get it.
thanks for any help!!

Use the Cauchy product formula,
$\displaystyle e^ae^b = \left( \sum_{n=0}^{\infty} \frac{a^n}{n!} \right) \left( \sum_{n=0}^{\infty}\frac{b^n}{n!} \right) = \sum_{n=0}^{\infty} c_n$
Where $\displaystyle c_n = \sum_{k=0}^n \frac{a^k}{k!}\cdot \frac{b^{n-k}}{(n-k)!} = \frac{1}{n!} \sum_{k=0}^n {n\choose k}a^kb^{n-k} = \frac{(a+b)^n}{n!}$
Therefore,
$\displaystyle e^ae^b = \sum_{n=0}^{\infty} \frac{(a+b)^n}{n!} = e^{a+b}$.
• Sep 15th 2008, 05:25 PM
Soroban
Hello, gidget!

Quote:

I need the proof for: .$\displaystyle e^a\!\cdot\!e^b\;=\;e^{a+b}$
using the power series expansion: .$\displaystyle e^z\:=\:1+z+\frac{z^2}{2!} + \frac{z^3}{3!} +\cdots$

We have: .$\displaystyle \begin{array}{cccccccc} e^a &=& 1 &+\; a & +\; \dfrac{a^2}{2!} & +\; \dfrac{a^3}{3!} & +\; \dfrac{a^4}{4!} & + \hdots \\ \\[-3mm] e^b & = & 1 & +\; b & +\; \dfrac{b^2}{2} &+\; \dfrac{b^3}{3!} &+\; \dfrac{b^4}{4!} &+ \hdots \\ \\[-3mm] \hline \end{array}$

Multiply: . $\displaystyle \begin{array}{cccccccc} e^a\!\cdot\!e^b &=& 1 & +\;a & +\;\dfrac{a^2}{2!} & +\;\dfrac{a^3}{3!} & +\;\dfrac{a^4}{4!} & + \hdots \\ \\[-3mm] & & & +\;b & + \;ab & +\; \dfrac{a^2b}{2!} & +\;\dfrac{a^3b}{3!} & + \hdots \\ \\[-3mm] & & & & + \;\dfrac{b^2}{2!} & +\;\dfrac{ab^2}{2!} & +\;\dfrac{a^2b^2}{2!2!} & +\hdots \\ \\[-3mm] & & & & & + \;\dfrac{b^3}{3!} & +\;\dfrac{ab^3}{3!} & + \hdots \\ \\[-2mm] \end{array}$
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle \begin{array}{cc}\dfrac{b^4}{4!} & + \hdots\end{array}$

Add down the columns . . .

$\displaystyle e^a\!\cdot\!e^b \;=\;1 + (a + b) + \frac{a^2+2ab + b^2}{2!} + \frac{a^3+3a^2b+3ab^2 + b^3}{3!}$.$\displaystyle +\:\frac{a^4+4a^3b + 6a^2b^2 + 4ab^3 + b^4}{4!} + \hdots$

$\displaystyle e^a\!\cdot\!e^b\;=\;1 + (ab) + \frac{(a+b)^2}{2!} + \frac{(a+b)^3}{3!} + \frac{(a+b)^4}{4!} + \hdots \;=\;e^{a+b}$