$\mathbb{R}$ is complete $\rightarrow$ $\mathbb{C}$ is complete In a book I am reading about functional analysis and metric spaces. The author goes through a long proof to show that $\mathbb{C}$ is complete, but at the end of the proof the author states that a simpler proof is to use the completenss of $\mathbb{R}$ to prove the completeness of $\mathbb{C}$, I have tried to look for this proof but couldn't find any. Is there a way to proof this?

Suppose that $z_n=x_n+i y_n$ with $x_n,y_n\in\mathbb R$  is a Cauchy sequence in $\mathbb C$. Since $|z|^2=|x|^2+|y|^2$ this is possible iff and only if $\{x_n\}$ and $\{y_n\}$ are Cauchy sequences in $\mathbb R$. In fact,
$$
|z_n-z_m|^2=|x_n-x_m|^2+|y_n-y_m|^2\leq \epsilon\Rightarrow |x_n-x_m|^2\leq \epsilon\textrm{ and } |y_n-y_m|^2\leq \epsilon
$$
and viceversa
$$
 |x_n-x_m|^2\leq \epsilon\textrm{ and } |y_n-y_m|^2\leq \epsilon \Rightarrow |z_n-z_m|^2 \leq 2\epsilon.
$$
 Since $\mathbb R$ is complete,
$$
\exists \bar x,\bar y\in \mathbb R:\quad \lim_n x_n=\bar x,\quad \lim_n y_n=\bar y
$$
This immediately implies
$$
z_n\to \bar x+ i\bar y=:\bar z
$$
Therefore each Cauchy sequence in $\mathbb C$ is convergent in $\mathbb C$.
