# Complex Sequence is Cauchy iff Convergent/Proof 1

## 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 $\mathcal N$ be the domain of $\sequence {z_n}$.

Let $x_n = \Re \paren {z_n}$ for every $n \in \mathcal N$.

Let $y_n = \Im \paren {z_n}$ for every $n \in \mathcal N$.

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

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

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

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

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

### Necessary Condition

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

We aim to prove that $\sequence {z_n}$ is convergent.

We find:

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

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

- $\implies \sequence {x_n}$ and $\sequence {y_n}$ are convergent by Real Sequence is Cauchy iff Convergent

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

$\Box$

### Sufficient Condition

Let $\sequence {z_n}$ be convergent.

We aim to prove that $\sequence {z_n}$ is a Cauchy sequence.

We find:

- $\sequence {z_n}$ is convergent

- $\implies \sequence {x_n}$ and $\sequence {y_n} $ are convergent by definition of convergent complex sequence

- $\implies \sequence {x_n}$ and $\sequence {y_n}$ are Cauchy sequences by Real Sequence is Cauchy iff Convergent

- $\implies \sequence {z_n}$ is a Cauchy sequence by Lemma

$\blacksquare$