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?

  • $\begingroup$ But $\Bbb C$ is $\Bbb R^2$ with the operations of complex addition and multiplication! $\endgroup$ – Pedro Tamaroff Sep 12 '13 at 17:07
  • $\begingroup$ As sets $\Bbb C = \Bbb R^2$. $\endgroup$ – Martín-Blas Pérez Pinilla Feb 1 '18 at 11:50

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.

  • 1
    $\begingroup$ Thanks for the answer. I was thinking, maybe he did this to avoid problems when changing coordinates? If $z=x+iy=re^{i\theta}$, then $u(z)=u(x+iy)=u(re^{i\theta})$, but $u(x,y) \neq u(r,\theta)$, strictly speaking (at some point he gives the C-R equations in polar form.) $\endgroup$ – Spine Feast Sep 12 '13 at 17:43
  • $\begingroup$ Yes, that's a good observation. $\endgroup$ – John Gowers Sep 12 '13 at 17:45

$\mathbb R^2$ 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.


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