Cauchy's Convergence Criterion/Complex Numbers/Lemma 1

Lemma for Complex Sequence is Cauchy iff Convergent
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 $\sequence {x_n}$ and $\sequence {y_n}$ are (real) Cauchy sequences.

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

This means that, for a given $\epsilon > 0$:


 * $\exists N: \forall m, n \in \mathcal N: m, n \ge N: \cmod {z_n - z_m} < \epsilon$

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.

Sufficient Condition
Let $\sequence {x_n}$ and $\sequence {y_n}$ be Cauchy sequences.

This means for $\sequence {x_n}$ that, for a given $\epsilon > 0$:
 * $\exists N_1: \forall m, n \in \mathcal N: m, n \ge N_1: \cmod {x_n - x_m} < \dfrac \epsilon 2$

Also, for $\sequence {y_n}$:


 * $\exists N_2: \forall m, n \in \mathcal N: m, n \ge N_2: \cmod {y_n - y_m} < \dfrac \epsilon 2$

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.