Why is the magnitude of this vector $\sqrt{14}$?

Question

I don't understand why, when normalizing the vector $(2,-3-i)$, we get $(2,-3-i)/\sqrt{14}$. When I find the magnitude of the vector, I do

$$=\sqrt{(2)^2+(-3-i)^2}=\sqrt{4+8+6i}=\sqrt{12+6i}$$

How am I supposed to get $\sqrt{14}$??

Answer 1

When you find the magnitude of a vector with complex components, you want the absolute square of each component, where you multiply it by its conjugate. so $$|(2,-3-i)|=\sqrt{(2)^2+(-3-i)(-3+i)}=\sqrt{4+10}=\sqrt{14}$$