# Proof

We have to show that the complex exponential function $$\exp:\mathbb C\to \mathbb C$$ is continuous on whole $$\mathbb C$$, formally for any $$a\in\mathbb C$$, we have $\lim_{x\to a}\exp(x)=\exp(a).$ Let $$(x_n)_{n\in\mathbb N}$$ be a convergent complex series with $$\lim_{n\to\infty} x_n=a$$. We have then $$\lim(x_n-a)=0$$. Together with the result $$\exp(0)=1$$ it follows $\lim_{n\to \infty}\exp(x_n-a)=1.$ Because of the non-zero property of the complex exponential function $$\exp(x)\neq 0$$ for all $$x\in\mathbb C$$, and because of the functional equation of the complex exponential function, we can conclude that $1=\lim_{n\to \infty}\exp(x_n-a)=\frac{\lim_{n\to \infty}\exp(x_n)}{\lim_{n\to \infty}\exp(a)}=\lim_{n\to \infty}\frac{\exp(x_n)}{\exp(a)}.$ $\exp(a)=\lim_{n\to \infty}\exp(x_n).$ In the last step we have used the formula for the quotient of convergent complex sequences.

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### References

#### Bibliography

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