Cauchy's Convergence Criterion/Complex Numbers/Lemma 1

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Lemma for Cauchy's Convergence Criterion on Complex Numbers

Let $\sequence {z_n}$ be a complex sequence.

Let $\NN$ be the domain of $\sequence {z_n}$.

Let $x_n = \map \Re {z_n}$ for every $n \in \NN$.

Let $y_n = \map \Im {z_n}$ for every $n \in \NN$.


Then $\sequence {z_n}$ is a (complex) Cauchy sequence if and only if $\sequence {x_n}$ and $\sequence {y_n}$ are (real) Cauchy sequences.


Proof

Necessary Condition

Let $\sequence {z_n}$ be a Cauchy sequence.

This means that, for a given $\epsilon > 0$:

$\exists N: \forall m, n \in \NN: m, n \ge N: \cmod {z_n - z_m} < \epsilon$


We have, for every $m, n \ge N$:

\(\ds \cmod {x_n − x_m}\) \(=\) \(\ds \cmod {\map \Re {z_n − z_m} }\)
\(\ds \) \(\le\) \(\ds \cmod {z_n − z_m}\) Modulus Larger than Real Part
\(\ds \) \(<\) \(\ds \epsilon\)

Thus $\sequence {x_n}$ is a Cauchy sequence by definition.


A similar argument shows that $\sequence {y_n}$ is a Cauchy sequence.

$\Box$


Sufficient Condition

Let $\sequence {x_n}$ and $\sequence {y_n}$ be Cauchy sequences.

This means for $\sequence {x_n}$ that, for a given $\epsilon > 0$:

$\exists N_1: \forall m, n \in \NN: m, n \ge N_1: \cmod {x_n - x_m} < \dfrac \epsilon 2$

Also, for $\sequence {y_n}$:

$\exists N_2: \forall m, n \in \NN: m, n \ge N_2: \cmod {y_n - y_m} < \dfrac \epsilon 2$


Let $N = \max \paren {N_1, N_2}$.

Let $i = \sqrt {-1}$ denote the imaginary unit.

We have, for every $m, n \ge N$:

\(\ds \cmod {z_n − z_m}\) \(=\) \(\ds \cmod {x_n + i y_n − \paren {x_m + i y_m} }\) as $z_n = \map \Re {z_n} + i \map \Im {z_n}$ and $z_m = \map \Re {z_m} + i \map \Im {z_m}$
\(\ds \) \(=\) \(\ds \cmod {x_n − x_m + i \paren {y_n − y_m} }\)
\(\ds \) \(\le\) \(\ds \cmod {x_n − x_m} + \cmod {i\paren {y_n − y_m} }\) Triangle Inequality for Complex Numbers
\(\ds \) \(=\) \(\ds \cmod {x_n − x_m} + \cmod {y_n − y_m}\) Definition of Complex Modulus
\(\ds \) \(<\) \(\ds \frac \epsilon 2 + \cmod {y_n − y_m}\) since $\cmod {x_n - x_m} < \dfrac \epsilon 2$
\(\ds \) \(<\) \(\ds \frac \epsilon 2 + \frac \epsilon 2\) since $\cmod {y_n - y_m} < \dfrac \epsilon 2$
\(\ds \) \(=\) \(\ds \epsilon\)

Thus $\sequence {z_n}$ is a Cauchy sequence by definition.

$\blacksquare$

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