Complex numbers as vectors I want to know the separate circumstances under which complex numbers can or cannot be treated as vectors as relates to the following specific item.  Most importantly, rather than telling me that the language I'm using is wrong, please understand that I am seeking better language by which to describe the issue in question.  The main point of this post is for me to discover better language by which to make the following distinction so please understand that I do not already know the words I am looking for.  
For a complex number $z=a+ib$ we have 
$$z^2=(a+ib)(a+ib).$$
However, for a complex vector we have
$$  \vec z^2= (a+ib)(a-ib).  $$
I have found material online saying that every complex number is a vector, and that it is not always a vector.  Please help me understand the distinction between the two modes of multiplication: scalar multiplication and vector multiplication.  What language is used to make the distinction between the two cases?  How do I know if $z^2$ is supposed to have the complex conjugate in it?  I suppose I should probably say something about the norm of the vector, but I am not sure what the words are.  Please provide the formalism which makes the distinction clear.  THANKS.
 A: No, $(a+bi)(a-bi)$ is not $\vec{z}^2$, it is $\|\vec{z}\|^2$.  There is no such thing as the square of a vector.  $\|\vec{z}\|$ is the (Euclidean) length of the vector $\vec{z}$, which is $\sqrt{a^2 + b^2}$, and this happens to be the same
as $\sqrt{(a+bi)(a-bi)} = \sqrt{z \overline{z}}$.
A: Complex numbers are vectors. They form what's known as a vector space over $\Bbb{R}$, a term that I'm not going to explain, but is a formal, rigorous, mathematical notion of a set of objects that can be sensibly considered "vectors". Indeed, it's not hard to see how they can be identified with vectors in $\Bbb{R}^2$: you can swap between Cartesian form $a + ib \in \Bbb{C}$ and the vector $(a, b) \in \Bbb{R}^2$.
Note that this preserves the vector operations of addition and scalar multiplication (in this case, multiplying by a real number), in the sense that
$$(a + ib) + (c + id) \mapsto (a, b) + (c, d)$$
and if $r \in \Bbb{R}$,
$$r(a + ib) \mapsto r(a, b).$$
Essentially what I'm saying is, complex numbers, considered as vectors, behave identically to $\Bbb{R}^2$, and anything you can do in $\Bbb{R}^2$, there will be a parallel version in $\Bbb{C}$.
But, complex numbers are more than simply vectors in $\Bbb{R}^2$. We can multiply these complex numbers together, unlike vectors in $\Bbb{R}^2$. We can do calculus with complex numbers (differentiation and integration), unlike $\Bbb{R}^2$ (actually, we can differentiate in $\Bbb{R}^2$, but it is a different notion to differentiation with complex numbers).
So, in short, complex numbers are more than just vectors. If it helps in a problem to think of them as vectors, then please do so. However, you'll soon find that not every complex numbers problem can be solved in this way.
A: Various authors may well use a different notation than I'm outlining below. I have written what is commonly used, what makes for a consistent article, and what is aligned with wikipedia.
For convenience in the following let $r\in\mathbb R,\,c=a+ib\in\mathbb C,\, (u,v)\in\mathbb R^2,\,(x,y)\in\mathbb R^2,\,\mathbf w=(u+iv)\in\mathbb C^1,\,\mathbf z=(x+iy)\in\mathbb C^1,$ and let $\mathbf z^*=(x-iy)$ represent the conjugate transpose.
Vector spaces and their scalar product
We are talking about the following vector spaces over fields with vector addition and scalar multiplication. They are defined using the usual addition and multiplication of the corresponding fields.


*

*$(\mathbb R^2,\mathbb R,+,\cdot)$ with scalar product $r\cdot(x,y)=(rx,ry)$.

*$(\mathbb C^1,\mathbb R,+,\cdot)$ with scalar product $r\cdot \mathbf z = r\cdot(x+iy) =rx+iry$.

*$(\mathbb C^1,\mathbb C,+,\cdot)$ with scalar product $c\cdot \mathbf z = (a+ib)\cdot(x+iy) = (ax-by)+i(ay+bx)$.


The notation $\mathbb C^1$ is used here to emphasize that it is a set of vectors.
The notation $\mathbf z$ is used here for a vector, although we can also write $z$ if there is no ambiguity.
Vector spaces 1 and 2 are isomorphic. They are commonly treated as if they are simply the same.
Note that the multiplication dot is commonly left out for convenience (e.g. $r\mathbf z$).  
Dot product of vectors
The dot product is the sum of the products of the corresponding entries of two sequences of numbers.
Various authors may well use a different convention, and for instance treat the dot product as an inner product or vice versa.
But strictly speaking, for the vectors in each of the cases, it should be:


*

*The dot product $(u,v)\cdot(x,y)=ux+vy$.

*The dot product $\mathbf w \cdot \mathbf z = (u+iv)\cdot(x+iy)=(ux-vy) + i(uy+vx)$.

*The dot product $\mathbf w \cdot \mathbf z = (u+iv)\cdot(x+iy)=(ux-vy) + i(uy+vx)$.


It is actually ambiguous that we use a dot for regular multiplication, for the scalar product, and for the dot product. But given the context in which they are used, there is usually no confusion.
Inner products
The usual inner product in each case is:


*

*$\langle (u,v),\,(x,y)\rangle = ux+vy \in\mathbb R$.

*$\langle \mathbf w,\mathbf z \rangle=ux+vy \in\mathbb R$.

*$\langle \mathbf w,\mathbf z \rangle=\mathbf z^*\mathbf w = (u+iv)(x-iy) = (ux+vy)+i(vx-uy) \in\mathbb C$.


In case 1 the dot product and the inner product are the same, and they are used interchangeably. We can't do that in cases 2 and 3 though.
It is rare to see the inner product of case 2. Usually case 3 is implied.
Just for fun, in Quantum Physics the notation $\langle \phi|\psi\rangle = |\phi\rangle^\dagger|\psi\rangle$ is used for the inner product of two quantum states or probability densities.
Vector norms
The usual vector norm in each case is:


*

*$\|(x,y)\| = \sqrt{x^2+y^2} \in\mathbb R$.

*$\|\mathbf z\| = \sqrt{\langle \mathbf z,\mathbf z\rangle} = \sqrt{x^2+y^2}\in\mathbb R$.

*$\|\mathbf z\| = \sqrt{\langle \mathbf z,\mathbf z\rangle} = \sqrt{x^2+y^2}\in\mathbb R$.


No confusion here. They are all exactly as we would expect.
