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 $m, n \ge N$:

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 $m, n \ge N$:

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

$\blacksquare$