Proof
(related to Proposition: Convergent Complex Sequences Vs. Convergent Real Sequences)
\("\Rightarrow"\)
- 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 \(\lim_{n\to\infty}c_n=c\).
∎
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References
Bibliography
- Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983