# Show that the set of complex numbers is complete metric space

I know that the set of complex number is a normed linear with norm $\|z\|=|z|$. The induced metric is $d(z,w)=|z-w|$ for complex $z$ and $w$. But I want to prove that the set is complete.Thanks for any help

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Do you know that $\mathbb{R}$ is complete? –  Najib Idrissi May 5 '12 at 10:14

## 3 Answers

HINT: Show that if $\langle x_n+iy_n:n\in\Bbb N\rangle$ is a Cauchy sequence in $\Bbb C$, then $\langle x_n:n\in\Bbb N\rangle$ and $\langle y_n:n\in\Bbb N\rangle$ are Cauchy sequences in $\Bbb R$ with the usual metric. Then use the fact that $\Bbb R$ is complete.

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If $(z_n) = (x_n) + i (y_n)$ is Cauchy in $\mathbb{C}$ then $(x_n)$ and $(y_n)$ are Cauchy in $\mathbb{R}.$ Since $\mathbb{R}$ is complete, $x_n\to x$ and $y_n \to y,$ so $z_n \to z = x+iy.$

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HINT: Prove that a finite product of complete metric spaces is complete and use $\Bbb C\simeq\Bbb R^2$.

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