$\mathbb{R}$ 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?
 A: 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 $\mathbb C$.
A: A sequence in $\Bbb C$ converges iff the real part and the imaginary part converge.
The same holds for cauchy sequences but both, the real part sequence and the imaginary part sequence are cauchy sequences in $\Bbb R$, so you can use completeness of $\Bbb R$.
A: Hint: $$|z_n-w|\to0\Leftrightarrow |\Re{z_n}-\Re{w}|\to0\text{ and }|\Im{z_n}-\Im{w}|\to0$$
