Why does path-independence of limits hold in the complex numbers but not in $\mathbb{R}^2$?

Question

In my studies of complex differentiation, I've come across the following paradox concerning real partial derivatives:

In proofs of the Cauchy-Riemann equations, the fact that the limit \begin{align*} \frac{df}{dz}\Bigg\vert_{z=z_0}=\lim_{\Delta z\rightarrow 0}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z} \end{align*}

converges to the same value, independent of how $\Delta z$ approaches $0$. Using this fact, the equality

\begin{align} \frac{\partial f}{\partial x}\Bigg\vert_{z=z_0}=\lim_{\Delta x\rightarrow 0}\frac{f(z_0+\Delta x)-f(z_0)}{\Delta x}=\lim_{i\Delta y\rightarrow 0}\frac{f(z_0+i\Delta y)-f(z_0)}{i\Delta y}=\frac{\partial f}{\partial y}\Bigg\vert_{z=z_0} \end{align}

is arrived at, and is used to prove that the Cauchy-Riemann equations hold wherever $f$ is analytic.

The question, then, is: why does this equality of partial derivatives with the standard derivative, as well as with each other, $(\frac{df}{dz}=\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}$) hold when considering differentiation in $\mathbb{C}$, but not when considering multivariable differentiation in $\mathbb{R}^2$? For clearly, a function with a domain $D\subset\mathbb{R}^2$ are differentiable at many points where yield their partial derivatives differ: $\frac{\partial f}{\partial x}\neq\frac{\partial f}{\partial y}$, which apparently cannot occur in the case of complex partial derivatives. I'm not even quite sure what the analogous quantity of $\frac{df}{dz}$ would be in the real case.

What accounts for the difference in these two cases? What attribute does the complex plane possess that is absent in the real plane?

Thanks very much for reading - any help would be appreciated.

Answer 1

More of a ramble than an answer, but may highlight one key difference from $\mathbb{R}^2$ which is that elements of $\mathbb{C}$ are scalars.

In both cases we have $f(z+h)- f(z) \approx f'(z)h $ - nothing distinguishing here.

Consider a rotation $\rho \in \mathbb{C}$, that is $|\rho|=1$.

Then $f(z+\rho h)- f(z)) \approx f'(z) \rho h = \rho f'(z) h \approx \rho (f(z+h)-f(z))$. That is, the effect of rotating the 'perturbation' $h$ is the same rotating the effect of the perturbation.

This does not hold in general in $\mathbb{R}^2 \to \mathbb{R}^2$ mappings. For the same to happen with a rotation $R$ would require $R f'(x) = f'(x) R$, and the $2\times 2$ matrices which commute with any rotation $R$ are of the form $\lambda \begin{bmatrix} c & s \\ -s & c \end{bmatrix}$, for real scalars satisfying $c^s+s^2 =1$. This has the same form as required by the Cauchy Riemann equations, $\begin{bmatrix} u_x & u_y \\ v_x & v_y \end{bmatrix} = \begin{bmatrix} u_x & u_y \\ -u_y & u_x \end{bmatrix}$.

Answer 2

$\newcommand{\dd}{\partial}$The calculation in the question loses a factor of $-i$ at the last equality; it should read: \begin{align*} \frac{\dd f}{\dd x}\bigg\vert_{z=z_{0}} &= \lim_{\Delta x\to 0}\frac{f(z_{0} + \Delta x) - f(z_{0})}{\Delta x} \\ &= \lim_{i\Delta y\to 0}\frac{f(z_{0} + i\Delta y) - f(z_{0})}{i\Delta y} \\ &= -i\frac{\dd f}{\dd y}\bigg\vert_{z=z_{0}}. \end{align*}

To see why, recall that by using notation for real partial derivatives we're identifying complex numbers $z = x + iy$ with Cartesian pairs $(x, y)$. Under this identification, we have \begin{align*} \lim_{i\Delta y\to 0}\frac{f(z_{0} + i\Delta y) - f(z_{0})}{i\Delta y} &= \frac{1}{i} \lim_{i\Delta y\to 0}\frac{f(z_{0} + i\Delta y) - f(z_{0})}{\Delta y} \\ &= -i \lim_{i\Delta y\to 0}\frac{f(x_{0}, y_{0} + \Delta y) - f(x_{0}, y_{0})}{\Delta y}\qquad\text{(n.b. $y_{0} + \Delta y$)} \\ &= -i\frac{\dd f}{\dd y}\bigg\vert_{(x,y)=(x_{0},y_{0})} \\ &= -i\frac{\dd f}{\dd y}\bigg\vert_{z=z_{0}}. \end{align*}