Intuitive explanation why complex differentiation is not linear over real and imaginary part

Question

Suppose we have a complex valued function $f = u + iv \colon U \to \mathbb C$ defined on an open subset $U$ of $\mathbb C$, which is holomorphic.

I was asked why it is wrong to use the linearity of the differentiation operator to write $$ \frac{d}{dz} f(z) = \frac{d}{dz} (u(z) + iv(z)) = \frac{d}{dz} u(z) + i \frac{d}{dz} v(z). $$ I explained this by proving that a real - valued function that is holomorphic must be constant, and so for $\frac{d}{dz} u$ and $\frac{d}{dz} v$ to exist $u$ and $v$ must be constant in $z$. However $f$ is not necessarily constant so applying the linearity in this way is invalid. I also showed that the real - and imaginary part functions $\Re()$ and $\Im()$ are not holomorphic.

Intuitively I think that real - valued and complex - valued functions are a different 'thing' and the $\frac{d}{dz}$ operator lives in the 'complex analysis world', so even though it is linear over complex functions it does not make sense to use it in the context of real - valued functions. However I was wondering whether there is a clearer or perhaps deeper explanation to this.

Answer 1

The explanation you write is a correct explanation of why you can't necessarily talk about $\frac{d}{dz}u(z)$ and $\frac{d}{dz}v(z)$ - the limits simply don't exist in general. But we have to be careful here. What we are saying is that the following limits (taken in the complex plane) do not exist: $$ \frac{d}{dz}u(z) := \mbox{lim}_{h\rightarrow 0}\frac{u(z+h) - u(z)}{h}\\ \\ \frac{d}{dz}v(z) := \mbox{lim}_{h\rightarrow 0}\frac{v(z+h) - v(z)}{h}\\ $$ Existence of these limits is a very strong constraint because it is requires the same value for the limit no matter which 'direction' $h$ approaches $0$ from.

On the other hand, what is often done in complex analysis is that the (partial) differential operators $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\overline{z}}$ are defined to be $\frac{1}{2}\left(\frac{\partial}{\partial x} + \frac{1}{i}\frac{\partial }{\partial y}\right)$ and $\frac{1}{2}\left(\frac{\partial}{\partial x} - \frac{1}{i}\frac{\partial }{\partial y}\right)$ respectively. With these definitions, it is quite immediate that $\frac{\partial}{\partial z}f(z) = \frac{\partial}{\partial z}u(z) + i\frac{\partial }{\partial z}v(z)$ (these operators are linear). By the Cauchy-Riemann equations, you find that $\frac{\partial}{\partial \overline{z}}f(z) = 0$ for $f(z)$ holomorphic. Indeed, this condition on $f$ is in fact equivalent to holomorphicity. In this case too, it is true that $\frac{\partial}{\partial z}f(z) = \frac{d}{dz}f(z)$. But we still do not have $\frac{\partial }{\partial z}u(z) = \frac{d}{dz}u(z)$ (resp. for $v$) because the latter simply may not exist.

So some care is warranted in working with all these operators in complex analysis.

As to your intuition about real- and complex-valued functions. They surely are different beings, but really, real-valued functions are, in particular, complex-valued functions because there is a natural embedding of $\mathbb{R}$ in $\mathbb{C}$ (at least, once $i$ is 'defined'). You just have to keep in mind that not all smooth complex-valued functions are actually holomorphic.

Addendum (to expand on response to Tanuj's comment on the OP):

The issue is not about linearity of the operator $\frac{d}{dz}$ over $\mathbb{C}$ versus over $\mathbb{R}$. This operator is already $\mathbb{C}$-linear and $\mathbb{C}$-linearity implies $\mathbb{R}$-linearity (the converse need not be true of course). The issue can instead be viewed as follows: think of the differential operators $\frac{d}{dz}$, $\frac{\partial }{\partial z}$ and $\frac{\partial}{\partial\overline{z}}$ as linear maps between function spaces (which they are): $$ \begin{align*} \frac{d}{dz} &: \mathscr{O}(U) \rightarrow \mathscr{O}(U)\\ \\ \frac{\partial}{\partial z},\frac{\partial}{\partial\overline{z}} &: \mathcal{C}^{\infty}(U,\mathbb{C}) \rightarrow \mathcal{C}^{\infty}(U, \mathbb{C}) \end{align*} $$ where $\mathscr{O}(U)$ denotes the space of holomorphic functions on $U$ and $\mathcal{C}^{\infty}(U,\mathbb{C})$ denotes the space of smooth complex-valued functions on $U$. We have the (linear) inclusions $\mathscr{O}(U) \hookrightarrow \mathcal{C}^{\infty}(U,\mathbb{C})$ and $\mathcal{C}^{\infty}(U,\mathbb{R}) \hookrightarrow \mathcal{C}^{\infty}(U, \mathbb{C})$ and the fact that the only holomorphic real-valued functions are the constants amounts to saying that $\mathscr{O}(U)\cap \mathcal{C}^{\infty}(U,\mathbb{R}) = \mathbb{R}$ (the intersection taking place in $\mathcal{C}^{\infty}(U,\mathbb{C})$ and where we think of $\mathbb{R}\hookrightarrow \mathbb{C} \hookrightarrow \mathcal{C}^{\infty}(U,\mathbb{C})$ as the natural inclusion of the constant functions).

Now, what I have said above (before the addendum) essentially amounts to the statements that

1. $\frac{d}{dz}$ cannot be extended to a map $\mathcal{C}^{\infty}(U,\mathbb{C}) \rightarrow \mathscr{O}(U)$ by applying the same limit definition to an arbitrary complex-valued smooth function (this is not to say it can't be extended in some other stranger way, which in fact it could)
2. the operators $\frac{\partial }{\partial z}$ and $\frac{\partial }{\partial \overline{z}}$ are indeed linear on all of $\mathcal{C}^{\infty}(U,\mathbb{C})$ and $\mbox{ker}\left(\frac{\partial}{\partial \overline{z}}\right) = \mathscr{O}(U)$
3. the operator $\frac{\partial}{\partial z}$ restricted to the subspace $\mathscr{O}(U)$ coincides with the composition of $\frac{d}{dz}$ with the inclusion $\mathscr{O}(U) \hookrightarrow \mathcal{C}^{\infty}(U,\mathbb{C})$ and this follows from the Cauchy-Riemann equations.

Answer 2

I would say that the formula is correct, but you need to specify that $u=u(z, \overline{z})$ and $v=v(z, \overline{z})$ and treat $z$ and $\overline{z}$ as two independent coordinates. You are indeed using complex coordinates to parametrize a real space: you are on $\mathbb{C}$ but you gave away the complex structure so you really are on $\mathbb{R}^2$. You need two independent coordinates to parametrize it.

For example, $f(z)=z$ gives $u=\frac{1}{2}(z+\overline{z})$ and $v=\frac{1}{2i}(z-\overline{z})$. The formula is now correct: \begin{equation} \begin{split} 1=\frac{dz}{dz} = \frac{\partial}{\partial z} \left( \frac{1}{2}(z+\overline{z})\right) + \frac{\partial}{\partial z} i\left(\frac{1}{2i}(z-\overline{z})\right) =1. \end{split} \end{equation}

See also Wirtinger derivatives.

Answer 3

The real and imaginary part are not analytic. The local Taylor expansion is

$u(z + h) = u(z) + ah + b\bar{h} + $ [sum of higher-degree polynomial functions of $(h,\bar{h})$].

The degree-1 approximation of $u(z)$ near $z_0$ is thus not of the form $u(z_0) +B (\Delta z)$ but $u(z_0) + B (\Delta z) + C(\overline{\Delta z})$.

In the typical situation where $C \neq 0$, the difference quotient $\frac{u(z + h) - u(z)}{h}$ will not converge to a limit as $h \to 0$, since it equals $(B+C\frac{\overline{h}}{h}+o(1))$ which converges only along (asymptotically) radial paths.

So the perturbation with respect to $z$ makes good sense, $du = u(z + dz) - u(z)$, but its description in terms of $dz$ is no longer consistent with the idea of approximate $\mathbb{C}$-linear dependence on $dz$ alone. Rather, you need to introduce $d\overline{z}$ and the derivative operators dual to it and $dz$.