# Proof

In the following $$(c_n)_{n\in\mathbb N}$$ is a complex sequence with $$c_n=a_n+i\cdot b_n$$ for some real numbers $$a_n,b_n$$. We will prove two inclusions:

### $$"\Rightarrow"$$

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

### $$"\Leftarrow"$$

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

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### References

#### Bibliography

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