# The length of a vector

I am struggling with a Linear Algebra problem that involves finding the length of a vector

$w_1 = (i, 1, 0) \in W$ a vector space over $\mathbb{C}$

So, the way I did it is like this:

$$\|w_1\| = \sqrt{i^2 + 1^2 + 0^2} = \sqrt{-1 + 1 + 0} = 0$$

Is this correct?

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That's a really short vector. You might want to check the definition of length in the complex case. – André Nicolas Apr 19 '12 at 0:41
BTW, $W$ is a vector space over $\mathbb{C}$. You could say $W$ is $\mathbb{C}^3$, and $\mathbb{C}^3$ is a vector space over $\mathbb{C}$. – Michael Hardy Apr 19 '12 at 0:46
Thanks for that. What's the LaTEX for the complex symbol? ... Nevermind, I copied it from the answer below. – adaptive Apr 19 '12 at 0:49

The length of a vector $v=(z_1, z_2, \ldots, z_n)\in \mathbb{C}^n$ is $||v||=\sqrt{z_1\overline{z_1}+\ldots + z_n\overline{z_n}}$, where $\overline z$ denotes the complex conjugate of $z$.
No. You have to multiply each component not by itself, but by its complex conjugate. A component is real if and only if it is its own complex conjugate, so you multiply $0$ by $0$, getting $0^2$, and $1$ by $1$, getting $1^2$. But $i$ is not real. Its complex conjugate is $-i$.