Does my proof of the dot product geometric and algebraic equivalence hold for R^2 when proven using complex numbers?

Question

I think I have come up with a proof of the equivalence between the algebraic and geometric definitions of the dot product in 2D using complex numbers, which goes as follows:

Let $z=a+bi=|z|e^{i\alpha}$ and $w=c+di=|w|e^{i\beta}$, then:

$$z\bar{w}= ac + bd + i(bc - ad)$$ $$\bar{z}w = ac + bd + i(ad - bc)$$

And by adding these and dividing by 2, the RHS becomes the algebraic definition of the dot product:

$$\frac{ac + bd + i(bc - ad) + ac + bd + i(ad - bc)}{2} = ac + bd$$

While the LHS becomes the geometric definition of the dot product:

$$\frac{z\bar{w} + \bar{z}w}{2} = \frac{|z|e^{i\alpha} \cdot |w|e^{i(-\beta)} + |z|e^{i(-\alpha)} \cdot |w|e^{i\beta}}{2}=|z||w|\frac{e^{i(\alpha-\beta)} + e^{i(-(\alpha-\beta))}}{2} = |z||w|cos(\alpha - \beta)$$

And thus: $$ac + bd = |z||w|cos(\alpha - \beta)$$

My question then is: given that my proof is done entirely with complex numbers does my proof imply the equivalence for vectors in $R^2$? Why/why not?
If not, would there be a way to make the proof for complex numbers imply the equivalence in $R^2$?

My thoughts: I have been looking at things like this, but I'm not really sure if $C$ being an isomorphism to $R^2$ is enough to make the proof hold for $R^2$ as well, or if it has to be a ring to hold since I don't have that much experience with either of them. If these are necessary concepts to understand why or why not it holds for $R^2$, then I would appreciate some links or other references to learn more about it.

Answer 1

There is no such thing as algebraic definition or geometric definition actually. The scalar product, or to be clear, the standard scalar product is the one you just wrote: $<z,w>=ac+bd$. Any other peculiarity holds true because the complex field is actually a nice structure which has the conjugacy of a vector, namely $\bar{z}$, so that you can derive some identities as the one you just found.

The identity you found is the very definition of cosine of an angle on a vector space, so that is nothing special either, although there are not mistakes in you computations.

In any case $\mathbb{C}$ and $\mathbb{R}^{2}$ are isomorphic both as fields and as vector spaces, and the isomorphism is actually the map which sends $z=x+iy\in\mathbb{C}$ to the couple $(x,y)\in\mathbb{R}^{2}$ so yeah they are equivalent, but that in no way depends on the dot product.