# 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 ProofOfAdditionFormulaOfExp 2013-03-22 16:32:03 2013-03-22 16:32:03 pahio (2872) pahio (2872) 4 pahio (2872) Proof msc 30D20 AdditionFormula AdditionFormulas addition formula of exponential function