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Math Help - [SOLVED] Complex exponential proof

  1. #1
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    [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!!
    Last edited by gidget; September 15th 2008 at 04:18 PM.
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  2. #2
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    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.
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  3. #3
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    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.
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  4. #4
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    Quote Originally Posted by icemanfan View Post
    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?

    Quote Originally Posted by gidget View Post
    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}.
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  5. #5
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    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}<br />
e^a &=& 1 &+\; a  & +\; \dfrac{a^2}{2!} & +\; \dfrac{a^3}{3!} & +\; \dfrac{a^4}{4!} & + \hdots \\ \\[-3mm]<br />
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}<br />
e^a\!\cdot\!e^b &=& 1 & +\;a & +\;\dfrac{a^2}{2!} & +\;\dfrac{a^3}{3!} & +\;\dfrac{a^4}{4!} & + \hdots \\ \\[-3mm]<br />
& & & +\;b & + \;ab & +\; \dfrac{a^2b}{2!} & +\;\dfrac{a^3b}{3!} & + \hdots \\ \\[-3mm]   <br />
& & & & + \;\dfrac{b^2}{2!} & +\;\dfrac{ab^2}{2!} & +\;\dfrac{a^2b^2}{2!2!} & +\hdots \\ \\[-3mm]<br />
& & & & & + \;\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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