[SOLVED] Complex exponential proof

**gidget** · September 15th 2008, 03:40 PM

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!!

**icemanfan** · September 15th 2008, 03:46 PM

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

**gidget** · September 15th 2008, 03:49 PM

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.

**ThePerfectHacker** · September 15th 2008, 04:49 PM

Originally Posted by icemanfan

$e^{z1}e^{z2} = e^{z1+z2}$ 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 $a^ba^c=a^{b+c}$ works for $a,b,c>0$ . But you cannot use this rule here because $z_1,z_2$ are complex numbers. Thus, would does it mean to raise an exponent?

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,
$e^ae^b = \left( \sum_{n=0}^{\infty} \frac{a^n}{n!} \right) \left( \sum_{n=0}^{\infty}\frac{b^n}{n!} \right) = \sum_{n=0}^{\infty} c_n$
Where $c_n = \sum_{k=0}^n \frac{a^k}{k!}\cdot \frac{b^{n-k}}{(n-k)!} = \frac{1}{n!} \sum_{k=0}^n {n\choose k}a^kb^{n-k} = \frac{(a+b)^n}{n!}$
Therefore,
$e^ae^b = \sum_{n=0}^{\infty} \frac{(a+b)^n}{n!} = e^{a+b}$ .

**Soroban** · September 15th 2008, 05:25 PM

Hello, gidget!

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

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

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

Add down the columns . . .

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

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

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