Is it valid to express a complex number as a vector?

[Mayhem]

...and is it ever useful?

An arbitrary complex number has the form $z = a + bi$ where $a, b \in \mathbb{R}$ and the dot product of two arbitrary vectors $\vec{v} = \binom{v_1}{v_2}$ and equivalently for vector $\vec{w}$ is $\vec{v} \cdot \vec{w} = v_1 w_1 + v_2 w_3$ Then the $z$ may be expressed as $z = a+bi = \vec{r} \cdot \vec{c} = \binom{a}{b} \cdot \binom{1}{i}$ where $\vec{r}$ denotes the real part and $\vec{c}$ the imaginary part.

Then in the complex plane, their composite vector $\vec{p}$ may be expressed using their magnitudes, giving $\vec{p} = \binom{||r||}{||c||} = \binom{\sqrt{a^2+b^2}}{\sqrt{2}i}$

Math seems valid unless I made a stupid mistake. Is this ever useful?

 

[DrClaude]

Expressing complex numbers as vectors is very common.

https://nrich.maths.org/2432

Not however that complex multiplication does not follow exactly vector multiplication, see https://geometricfunctions.org/curriculum/complex-plane/vector-products/index.shtml

 

[fresh_42]

...and is it ever useful?

An arbitrary complex number has the form $z = a + bi$ where $a, b \in \mathbb{R}$ and the dot product of two arbitrary vectors $\vec{v} = \binom{v_1}{v_2}$ and equivalently for vector $\vec{w}$ is $\vec{v} \cdot \vec{w} = v_1 w_1 + v_2 w_3$ Then the $z$ may be expressed as $z = a+bi = \vec{r} \cdot \vec{c} = \binom{a}{b} \cdot \binom{1}{i}$ where $\vec{r}$ denotes the real part and $\vec{c}$ the imaginary part.

Then in the complex plane, their composite vector $\vec{p}$ may be expressed using their magnitudes, giving $\vec{p} = \binom{||r||}{||c||} = \binom{\sqrt{a^2+b^2}}{\sqrt{2}i}$

Math seems valid unless I made a stupid mistake. Is this ever useful?

It depends on what you want to do.

The complex numbers are a two-dimensional real vector space $V$: $a+ib \mapsto (a,b).$

If we want to keep the multiplication, we have now first to define a two-dimensional, real algebra by
$$
(a\, , \,b) \circ (c\, , \,d) := (ac-bd\, , \,ad+bc)
$$
which is a multiplication that seems weird from the point of view of a vector space. Furthermore, all multiplication rules have to be proven again. Being a real vector space, we have an additional inner product defined by $(a,b)\cdot (c,d)=ab+cd \in \mathbb{R}$ which is not directly related to complex numbers and can be a source of confusion.

So, yes, you can consider the complex numbers as a real vector space $V$, but even the natural process of complexification
$$
V_\mathbb{C}=V\otimes_\mathbb{R} \mathbb{C}
$$
will end up in a total mess if you aren't very cautious; let alone complex calculus!

The complex numbers are usually only seen as a real vector space if we want to draw them. This is in my opinion already misleading and creates misconceptions. There is little advantage in such a concept except for being an example of a two-dimensional, real vector space. This advantage will be lost the moment you want to use them as a field and do analysis.

 

[FactChecker]

The complex numbers are usually only seen as a real vector space if we want to draw them. This is in my opinion already misleading and creates misconceptions. There is little advantage in such a concept except for being an example of a two-dimensional, real vector space. This advantage will be lost the moment you want to use them as a field and do analysis.

I tend to disagree with this. The geometry of complex multiplication, analysis, conformal mapping, etc., is beautifully seen in two dimensions.

 

[fresh_42]

I tend to disagree with this. The geometry of complex multiplication, analysis, conformal mapping, etc., is beautifully seen in two dimensions.

This likely depends on what should be visualized. I think of branching, Cauchy, Stokes, and that even $f(z)=z^2$ cannot be drawn anymore.

 

[PeroK]

This likely depends on what should be visualized. I think of branching, Cauchy, Stokes, and that even $f(z)=z^2$ cannot be drawn anymore.

You're not a fan of the argand diagram?

Using the geometry of the complex plane is invaluable to the average student.

 

[fresh_42]

You're not a fan of the argand diagram?

No.

Using the geometry of the complex plane is invaluable to the average student.

I think the field character gets lost. Even $\mathbb{R}[x]/(x^2+1)$ looks more like a vector space than a field. But the field property is essential for analysis.

However, I admit that this is a personal opinion. I haven't tried to figure out whether my lack of intuition is due to the vector space image or only a matter of the fact that complex numbers cannot be totally ordered.

 

[FactChecker]

This likely depends on what should be visualized. I think of branching, Cauchy, Stokes, and that even $f(z)=z^2$ cannot be drawn anymore.

I admit that I have to visualize a third dimension for the spirals of a Riemann surface, but ---
OUCH! I think I sprained a brain cell visualizing that. ;-)

 

[fresh_42]

I admit that I have to visualize a third dimension for the spirals of a Riemann surface, but ---
OUCH! I think I sprained a brain cell visualizing that. ;-)

My professor liked to use the following image for branching:

It helped a bit.

Edit: But the vector space is lost.