proof of addition formula of exp

The addition formula

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.