When we consider $\mathbb R^2$ but use the complex coordinate $z=x+iy,\bar z=x-iy$, why does $i$ behave like a scalar?
The background is that, simply from the definition of a complex coordinate that $i$ is nothing but a symbol. However, in actual computations the $i$ behaves like a number with nice properties.
I had this problem when I was trying to show $$dz\otimes d\bar z=dx\otimes dx+dy\otimes dy$$ By definition, $$dz\otimes d\bar z=(dx+idy)\otimes(dx-idy)\\ =dx\otimes dx+dx\otimes(idy)-dx\otimes (idy)-(idy)\otimes(idy)\\ =dx\otimes dx+idx\otimes dy-idx\otimes idy-i^2dy\otimes dy \\=dx\otimes dx+dy\otimes dy\\ $$ But wait, in the third line $i$ is moved to the front of the tensor, i.e. behaves like a scalar. However, these tensors belong to a vector field over $\mathbb R$. And $i$ is not real, hence I don't see a reason allowing us to move $i$ outside.
So what is the justification for this?