(z_n) is Cauchy IFF (Re(z_n)) and (Im(z_n)) are Cauchy

# A Sequence of Complex Numbers is Cauchy IFF The Real Part Sequence and Imaginary Part Sequence are Cauchy

 Proposition 2: If $(z_n)_{n=1}^{\infty} = (x_n + iy_n)_{n=1}^{\infty}$ is a sequence of complex numbers then $(z_n)$ is Cauchy if and only if both $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ are Cauchy sequences of real numbers.
• Proof: $\Rightarrow$ Suppose that $(z_n)$ is Cauchy. Let $\epsilon > 0$ be given. Then there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $|z_m - z_n| < \epsilon$. Then if $n \geq N$ we have that:
(1)
\begin{align} \quad |x_m - x_n| = \sqrt{(x_m - x_n)^2} \leq \sqrt{(x_m - x_n)^2 + (y_m - y_n)^2} = |z_m - z_n| < \epsilon \end{align}
• And:
(2)
\begin{align} \quad |y_m - y_n| = \sqrt{(y_m - y_n)^2} \leq \sqrt{(x_m - x_n)^2 + (y_m - y_n)^2} = |z_m - z_n| < \epsilon \end{align}
• So $(x_n)$ and $(y_n)$ are both Cauchy sequences of real numbers.
• $\Leftarrow$ Suppose that both $(x_n)$ and $(y_n)$ are Cauchy. Let $\epsilon > 0$ be given. Since $(x_n)$ is Cauchy there exists an $N_1 \in \mathbb{N}$ such that if $m, n \geq N_1$ then $|x_m - x_n| < \epsilon/2$. Similarly, since $(y_n)$ is Cauchy there exists an $N_2 \in \mathbb{N}$ such that if $m, n \geq N_2$ then $|y_m - y_n| < \epsilon/2$. Let $N = \max \{N_1, N_2\}$. Then if $m, n \geq N$ we have that:
(3)
\begin{align} \quad |z_m - z_n| = |(x_m + iy_m) - (x_n + iy_n)| = |(x_m - x_n) + i(y_m - y_n)| \leq |x_m - x_n| + |y_m - y_n| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
• So $(z_n)$ is Cauchy. $\blacksquare$