# Cauchy's Convergence Criterion/Complex Numbers/Lemma 1

## 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.

$\blacksquare$