# Cauchy's Convergence Criterion/Complex Numbers/Proof 2

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## Theorem

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

Then $\sequence {z_n}$ is a Cauchy sequence if and only if it is convergent.

## Proof

### Lemma

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.

$\Box$

Let $\sequence {x_n}$ be a real sequence where:

- $x_n = \map \Re {z_n}$ for every $n$
- $\map \Re {z_n}$ is the real part of $z_n$

Let $\sequence {y_n}$ be a real sequence where

- $y_n = \map \Im {z_n}$ for every $n$
- $\map \Im {z_n}$ is the imaginary part of $z_n$

We find:

- $\sequence {z_n}$ is a Cauchy sequence

- $\leadstoandfrom \sequence {x_n}$ and $\sequence {y_n}$ are Cauchy sequences by Lemma

- $\leadstoandfrom \sequence {x_n}$ and $\sequence {y_n}$ are convergent by Cauchy's Convergence Criterion on Real Numbers

- $\leadstoandfrom \sequence {z_n}$ is convergent by definition of convergent complex sequence

$\blacksquare$