Proof

(related to Proposition: Functional Equation of the Complex Exponential Function)

Let \(x,y\in\mathbb R\) be complex numbers. By definition of the complex exponential function,

\[\exp(x)=\sum_{n=0}^\infty\frac{x^n}{n!}\quad\text{ and }\quad\exp(y)=\sum_{n=0}^\infty\frac{y^n}{n!}\]

are absolutely convergent complex series for all \(x,y\in\mathbb C\). Therefore, their Cauchy product. \[\sum_{n=0}^\infty c_n=\left(\sum_{n=0}^\infty\frac{x^n}{n{!}}\right)\cdot\left(\sum_{n=0}^\infty\frac{y^n}{n{!}}\right)\] is also an absolutely convergent complex series. Moreover, applying the binomial theorem, we get \[c_n:=\sum_{k=0}^n\frac{x^{n-k}}{(n-k)! }\cdot\frac{y^k}{k! }=\frac1{n!}\sum_{k=0}^n\binom nk x^{n-k}y^k=\frac{(x+y)^n}{n! }.\] It follows \[\exp(x+y)=\sum_{n=0}^\infty\frac{(x+y)^n}{n! }=\left(\sum_{n=0}^\infty\frac{x^n}{n{ ! }}\right)\cdot\left(\sum_{n=0}^\infty\frac{y^n}{n{! }}\right)=\exp(x)\cdot \exp(y).\]


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References

Bibliography

  1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983