Complex Sequence is Cauchy iff Convergent/Proof 1

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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 $\mathcal N$ 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

$\blacksquare$