proof of addition formula of exp

e^{z_{1}+z_{2}}=e^{z_{1}}e^{z_{2}}

of the complex exponential function may be proven by applying Cauchy multiplication rule to the Taylor series expansions (http://planetmath.org/TaylorSeries) of the right side factors (http://planetmath.org/Product). We present a proof which is based on the derivative of the exponential function.

Let $a$ be a complex constant. Denote $e^{z}=w(z)$ . Then $w^{\prime}(z)\equiv w(z)$ . Using the product rule and the chain rule we calculate:

\frac{d}{dz_{1}}[w(z_{1})w(a-z_{1})]=w^{\prime}(z_{1})w(a-z_{1})+w(z_{1})w^{% \prime}(a-z_{1})(-1)=e^{z_{1}}e^{a-z_{1}}-e^{z_{1}}e^{a-z_{1}}\equiv 0

Thus we see that the product $w(z_{1})w(a-z_{1})=e^{z_{1}}e^{a-z_{1}}$ must be a constant $A$ . If we choose specially $z_{1}=0$ , we obtain:

A=w(0)w(a-0)=e^{0}e^{a}=e^{a}

Therefore

e^{z_{1}}e^{a-z_{1}}\equiv e^{a}.

If we denote $a-z_{1}:=z_{2}$ , the preceding equation reads $e^{z_{1}}e^{z_{2}}=e^{z_{1}+z_{2}}$ . Q.E.D.

Title	proof of addition formula of exp
Canonical name	ProofOfAdditionFormulaOfExp
Date of creation	2013-03-22 16:32:03
Last modified on	2013-03-22 16:32:03
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	4
Author	pahio (2872)
Entry type	Proof
Classification	msc 30D20
Related topic	AdditionFormula
Related topic	AdditionFormulas
Defines	addition formula of exponential function