Cauchy's Convergence Criterion/Complex Numbers/Lemma 1

Lemma for Complex Sequence is Cauchy iff Convergent
Let $\left \langle {z_n} \right \rangle$ be a complex sequence.

For all $i \in \N$, let $z_i = x_i + i y_i$.

Then $\left \langle {z_n} \right \rangle$ is a (complex) Cauchy sequence $\left \langle {x_n} \right \rangle$ and $\left \langle {y_n} \right \rangle$ are themselves (real) Cauchy sequences.

Necessary Condition
Let $\left \langle {z_n} \right \rangle$ be a Cauchy sequence.

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


 * $\exists N: \forall m, n \in \N: m, n \ge N: \left|{z_n - z_m}\right| < \epsilon$

where $\left|{z_n - z_m}\right|$ denotes the

We have, for $m, n \ge N$:

Thus, by definition, $\left\langle{x_n}\right\rangle$ is a Cauchy sequence.

A similar argument shows that $\left \langle {y_n} \right \rangle$ is a Cauchy sequence.

Sufficient Condition
Let $\left \langle {x_n} \right \rangle$ and $\left \langle {y_n} \right \rangle$ be Cauchy sequences.

This means for $\left \langle {x_n} \right \rangle$ that, for a given $\epsilon > 0$:
 * $\exists N_1: \forall m, n \in \N: m, n \ge N_1: \left|{x_n - x_m}\right| < \dfrac \epsilon 2$

Also, for $\left \langle {y_n} \right \rangle$:


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

Let $N = \max \left({N_1, N_2}\right)$.

We have, for $m, n \ge N$:

Thus, by definition, $\left \langle {z_n} \right \rangle$ is a Cauchy sequence.