Cauchy Sequences of Complex Numbers

# Cauchy Sequences of Complex Numbers

Recall from the Sequences of Complex Numbers page that a sequence of complex numbers $(z_n)_{n=1}^{\infty}$ is simply just an infinite order list of complex numbers.

We will now state an important type of sequence of complex numbers.

Definition: A sequence of complex numbers $(z_n)_{n=1}^{\infty}$ is a Cauchy Sequence of Complex Numbers if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $\mid z_m - z_n \mid < \epsilon$. |

It is a basic property that every Cauchy sequence of real numbers converges and hence the set of real numbers is said to be complete (a complete metric space to be more precise). Fortunately, every Cauchy sequence of complex numbers also converges. To establish this, we will need to prove the following result which states that a sequence of complex numbers converges if and only if both its real and imaginary parts converge as real sequences.

Lemma 1: Let $(z_n)_{n=1}^{\infty} = (x_n + i y_n)_{n=1}^{\infty}$ be a sequence of complex numbers. Then $(z_n)_{n=1}^{\infty}$ converges to $Z = X + Yi$ if and only if $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ converge as real sequences to $X$ and $Y$ respectively. |

**Proof:**$\Rightarrow$ Suppose that $(x_n + iy_n)_{n=1}^{\infty}$ converges to $X + Yi$. Then for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:

\begin{align} \quad \mid z_n - Z \mid = \mid x_n + iy_n - (X + Yi) \mid = \mid x_n - X + i(y_n - Y) \mid = \sqrt{(x_n - X)^2 + (y_n - Y)^2} < \epsilon \end{align}

- For this $N$ we have that if $n \geq N$ then:

\begin{align} \quad \mid x_n - X \mid = \sqrt{(x_n - X)^2} < \sqrt{(x_n - X) + (y_n - Y)^2} < \epsilon \end{align}

(3)
\begin{align} \quad \mid y_n - Y \mid = \sqrt{(y_n - Y)^2} < \sqrt{(x_n - X) + (y_n - Y)^2} < \epsilon \end{align}

- Therefore $(x_n)_{n=1}^{\infty}$ converges to $X$ and $(y_n)_{n=1}^{\infty}$ converges to $Y$.

- $\Leftarrow$ Conversely, suppose that $(x_n)_{n=1}^{\infty}$ converges to $X$ and $(y_n)_{n=1}^{\infty}$ converges to $Y$]. Let $\epsilon > 0$ be given. Then there exists $N_1, N_2 \in \mathbb{N}$ such that if $n \geq N_1$ we have that $\mid x_n - X \mid < \frac{\epsilon}{2}$ and if $n \geq N_2$ we have that $\mid y_n - Y \mid < \frac{\epsilon}{2}$. Let $N = \max \{ N_1, N_2 \}$. Then if $n \geq N$ we have that:

\begin{align} \quad \mid z_n - Z \mid = \mid (x_n + y_ni) - (X + Yi) \mid = \mid (x_n - X) + i(y_n - Y) \mid \leq \mid x_n - X \mid + \mid y_n - Y \mid < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

- Therefore $(z_n)_{n=1}^{\infty}$ converges to $Z$. $\blacksquare$

Lemma 2: If $(z_n)_{n=1}^{\infty} = (x_n + iy_n)_{n=1}^{\infty}$ is a Cauchy sequence of complex numbers then $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ are Cauchy sequences of real numbers. |

**Proof:**Suppose that $(z_n)_{n=1}^{\infty}$ is a Cauchy sequence of complex numbers. Then for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ we have that:

\begin{align} \quad \mid z_m - z_n \mid = \mid (x_m - x_n) + (y_m - y_n)i \mid = \sqrt{(x_m - x_n)^2 + (y_m - y_n)^2} < \epsilon \end{align}

- So for this $N$, if $m, n \geq N$ we have that:

\begin{align} \quad \mid x_m - x_n \mid = \sqrt{(x_m - x_n)^2} < \sqrt{(x_m - x_n)^2 + (y_m - y_n)^2} < \epsilon \end{align}

(7)
\begin{align} \quad \mid y_m - y_n \mid = \sqrt{(y_m - y_n)^2} < \sqrt{(x_m - x_n)^2 + (y_m - y_n)^2} < \epsilon \end{align}

- So $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ are Cauchy sequences of real numbers. $\blacksquare$.

Theorem 3: Every Cauchy sequence of complex numbers converges. In other words, the set of complex numbers is a complete metric space. |

**Proof:**Let $(z_n)_{n=1}^{\infty}$ be a Cauchy sequence of complex numbers. Then by Lemma 2, if $(z_n)_{n=1}^{\infty} = (x_n + y_ni)_{n=1}^{\infty}$ then $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ both converge as sequences of real numbers. But by Lemma 1 this means that $(z_n)_{n=1}^{\infty}$ converges. $\blacksquare$