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

Mayhem
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...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?
 
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Mayhem said:
...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.
 
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fresh_42 said:
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.
 
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FactChecker said:
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.
 
fresh_42 said:
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.
 
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PeroK said:
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.
 
fresh_42 said:
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. ;-)
 
FactChecker said:
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:

1656424559086.jpeg


It helped a bit.

Edit: But the vector space is lost.
 
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