normalize a vector in $\mathbb C^3$ - a very basic question

Question

I think I forgot a bit previous-year Linear Algebra, so I have a very basic question to you.

Given the following question:

Normalize the following vector: $v \in \mathbb {C^3}, \space v = i, -i, -2-2i$ using the standard inner product.

As far as I know, the standard inner product for $\mathbb C^n$ is $\langle {f, g} \rangle = \sum_{i=1}^n f_i \overline{g_i}$.

$v \in \mathbb C^3 \implies n = 3$.

The norm vector of $v$ is $\frac{v}{||v||}$.

$v$ is given, than I'm trying to find $||v||$:

$$||v|| = \sqrt{\langle v, v \rangle} = \sqrt{\sum_{i=1}^3 v_i\overline{v_i}} = \sqrt{\sum_{i=1}^3|v_i|}$$

I have bit forgotten what does $v_i$ means.

in this example, $v_1 = i, v_2 = -i, v_3 = -2-2i$?

If so, the answer is simply:

$$ |v| = \sqrt{|i| + |-i| + |-2-2i} = \sqrt{\sqrt{1} + \sqrt{1} + \sqrt{(-2)^2 + (-2)^2}} = \sqrt{2 + \sqrt{8}} \implies norm(v) = \frac{i, -i, -2-2i}{}\sqrt{2 + \sqrt{8}}?$$

Thanks in advance!!!

Answer 1

You have a mistake in your definition of $||v||$: $v_i \overline{v}_i = |v_i|^{\color{red} 2}$. The rest seems to be okay (not the numbers because of the wrong definition).