Proof of The Complex Numbers are Complete in the Modulus Metric

Throughout, claims 1-9 refer to Properties of Complex Conjugation and Modulus, and $\operatorname{Re}$, $\operatorname{Im}$ are the real and imaginary parts. For a real number $a$, its modulus as a complex number equals $a$ if $0\le a$ and $-a$ otherwise, by claim 8, so it agrees with the absolute value used in the definition of a Cauchy sequence of real numbers and in the definition of the limit of a real sequence; we write $|a|$ for it in either reading.

Step 1 (two estimates). Let $w\in\mathbb{C}$. By claim 6, $\operatorname{Re}w\le|w|$ and $-\operatorname{Re}w\le|w|$, so $|\operatorname{Re}w|\le|w|$; likewise $|\operatorname{Im}w|\le|w|$. Next, $|i|$ is the nonnegative real number with $|i|^{2}=0^{2}+1^{2}=1$, since $i$ has real part $0$ and imaginary part $1$; as $1$ is also nonnegative with square $1$, Existence and Uniqueness of the Nonnegative Square Root gives $|i|=1$. Hence, writing $w=\operatorname{Re}w+(\operatorname{Im}w)i$ and using the triangle inequality of claim 7 and the multiplicativity of claim 4,

Step 2 (the component sequences are Cauchy). Let $(z_{m})$ be a sequence of complex numbers that is Cauchy in $(\mathbb{C},d_{\mathbb{C}})$, and put $x_{m}=\operatorname{Re}z_{m}$ and $y_{m}=\operatorname{Im}z_{m}$. By claim 4 of Canonical Form and Arithmetic of Complex Numbers, $\operatorname{Re}(z_{m}-z_{l})=x_{m}-x_{l}$ and $\operatorname{Im}(z_{m}-z_{l})=y_{m}-y_{l}$. So by Step 1,

Given a real $\varepsilon>0$, choose $N$ with $d_{\mathbb{C}}(z_{m},z_{l})<\varepsilon$ for all $m,l\ge N$; then $|x_{m}-x_{l}|<\varepsilon$ and $|y_{m}-y_{l}|<\varepsilon$ for all $m,l\ge N$. Hence $(x_{m})$ and $(y_{m})$ are Cauchy sequences of real numbers.

Step 3 (passage to the limit). By Every Cauchy Sequence of Real Numbers Converges there are real numbers $x$ and $y$ with $x_{m}\to x$ and $y_{m}\to y$. Put $z=x+yi$. By claim 4 of Canonical Form and Arithmetic of Complex Numbers, $z_{m}-z$ has real part $x_{m}-x$ and imaginary part $y_{m}-y$, so Step 1 gives

Let $\varepsilon>0$ be real. Since $0<2$ and $2\neq0$ in the ordered field of real numbers, $\varepsilon/2$ is a positive real number. Choose $N_{1}$ with $|x_{m}-x|<\varepsilon/2$ for $m\ge N_{1}$ and $N_{2}$ with $|y_{m}-y|<\varepsilon/2$ for $m\ge N_{2}$, and let $N$ be the larger of $N_{1}$ and $N_{2}$. For every $m\ge N$,

Thus $(z_{m})$ converges to $z$ in $(\mathbb{C},d_{\mathbb{C}})$.

Since every Cauchy sequence in $(\mathbb{C},d_{\mathbb{C}})$ converges to a point of $\mathbb{C}$, the metric space $(\mathbb{C},d_{\mathbb{C}})$ is complete in the sense of Complete Metric Space.