# Thread: [SOLVED] Complex exponential proof

1. ## [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!!

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

3. 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.

4. 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?

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,
$\displaystyle e^ae^b = \sum_{n=0}^{\infty} \frac{(a+b)^n}{n!} = e^{a+b}$.

5. Hello, gidget!

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 . . .

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

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