# Proof

### $$"\Rightarrow"$$

• Assume, $$(c_n)_{n\in\mathbb N}$$ is a convergent complex sequence with $$\lim_{n\to\infty}c_n=c$$.
• Then, for each $$\epsilon > 0$$, there exists an $$N\in\mathbb N$$ with $| c_n-c | < \epsilon\quad\quad \text{ for all }n\ge N.$
• Because for every complex number $$x\in\mathbb C$$, we have $$|\Re(x)|\le |x|$$ and $$|\Im(x)|\le |x|$$ by the definition complex absolute value, it follows for all $$n\ge N$$: $\begin{array}{c} | a_n-a |=|\Re(c_n -c)|\le | c_n-c | < \epsilon\\ | b_n-b |=|\Im(c_n -c)|\le | c_n-c | < \epsilon\\ \end{array}$
• Thus, both real sequences $$(a_n)_{n\in\mathbb N}$$ and $$(b_n)_{n\in\mathbb N}$$ are convergent real sequences with $$\lim_{n\to\infty}a_n = \Re( c)=a$$ and $$\lim_{n\to\infty}b_n = \Im ( c)=b$$.

### $$"\Leftarrow"$$

• Assume, the real sequences $$(a_n)_{n\in\mathbb N}$$ and $$(b_n)_{n\in\mathbb N}$$ are convergent real sequences with $$\lim_{n\to\infty}a_n =a$$ and $$\lim_{n\to\infty}b_n=b$$.
• Then, for each $$\epsilon > 0$$ there exist indices $$N_a,N_b\in\mathbb N$$ with $| a_n-a | < \frac\epsilon2\quad\quad \text{ for all }n\ge N_a$ and $| b_n-b | < \frac\epsilon2\quad\quad \text{ for all }n\ge N_b.$
• Set $$c:=a+ib$$ and set $$N:=\max(N_a,N_b)$$.
• Then, for all $$n\ge N$$, it follows from the triangle inequality for the distance in the metric space of complex numbers that \begin{align} | c_n-c | &= |a_n+ib_n-a-ib| \nonumber\\ &= |(a_n-a)+i(b_n-b)| \nonumber\\ &\le |a_n-a|+|b_n-b|\nonumber\\ & < \frac\epsilon2+\frac\epsilon2\nonumber\\ &=\epsilon.\nonumber\end{align}
• Therefore, $$(c_n)_{n\in\mathbb N}$$ is a convergent complex sequence with $$\lim_{n\to\infty}c_n=c$$.

Github: ### References

#### Bibliography

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