A question on notation: $u(x,y)$ vs. $u(x+iy)$. In Tao's complex analysis notes, he uses the convention that $$f(z)=u(z)+iv(z)= u(x+iy)+iv(x+iy)$$
implying that the functions are $$u,v:\mathbb{C} \to \mathbb{R}.$$ But in other places I see the convention $$f(z)=u(x,y)+iv(x,y),$$ which implies that they functions are $$u,v:\mathbb{R}^2 \to \mathbb{R}.$$
Does this matter at all? It seems to me that it should, after all there's quite a few differences between the structures of $\mathbb{R}^2$ and $\mathbb{C}$, even though they are similar?
 A: There isn't really any difference in the structure of $\mathbb R^2$ and $\mathbb C$: $\mathbb C$ is just $\mathbb R^2$ with a multiplication operation put on it.  
I'd prefer to see $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$, which means that expressions like $\frac{\partial u}{\partial x}$ make more sense.  In fact, I'd even go so far as to say that it was a mistake on Tao's part to write $f(z)=u(x+iy)+v(x+iy)$, as it makes things a bit less clear.  On the other hand, I'm sure you know what he's talking about - it's just another convention and you can treat it as if it said $u(x,y)+v(x,y)$ anyway.  
In fact, even though Tao's $u$ is a function $\mathbb C\to\mathbb R$, there's only one canonical way to turn it into a function $\mathbb R^2\to\mathbb R$ (i.e., there's only one function $\hat u:\mathbb R^2\to\mathbb R$ such that $\hat u(x,y)=u(x+iy)$, so it's not losing any information to write things his way.
A: $\mathbb R^2$ is two-dimensional space over field of real numbers, but $\mathbb C$ can be considered either as two-dimensional space over field of real numbers or one-dimensional space over field of complex numbers.
If one considers $\mathbb C$ in the first case, it is easy to see that $\mathbb C=\mathbb R^2$ as a linear or vector space
In this case the elements of the space are pairs $(x,y)$ with operations $$(x,y)+(z,t)=(x+z,y+t)\\\lambda(x,y)=(\lambda x, \lambda y) \, \forall\lambda\in\mathbb R$$
Should we consider $\mathbb C$ in the second case, the spaces are not isomorphic, since the elements of $\mathbb C$ are numbers $z=x+iy$ with usual operations of sum and product. 
