0
$\begingroup$

I understand from this question that you can treat $\frac{\partial}{\partial \overline{z}} f$ as taking the two "conjugate conformal" non-orientation-conserving parts from $f$. And I get that if you have $f(z+h)=f(z)+h f'(z)+o(h)$, this implies that orientation is preserved. This is very interesting and enlightening, but I don't like that to come up with this you take partial derivatives while treating f as a function $\mathbb{R}^2\to \mathbb{R}^2$. Is there any way to phrase the Cauchy-Riemann equation(s) $\frac{\partial}{\partial \overline{z}} f=0$ without making appeals to multivariable calculus? Or is the best I can do simply the statement that $f'$ exists?

$\endgroup$
  • $\begingroup$ Don't partial derivatives per se belong to multivariate calculus and hence make your request futile? $\endgroup$ – Hagen von Eitzen May 30 '14 at 8:55
  • $\begingroup$ @HagenvonEitzen With a proper interpretation of $f$ as a function of multiple complex variables, no. $\endgroup$ – user18862 May 30 '14 at 9:50
  • $\begingroup$ I think this should always be phrased in terms of multivariable calculus, or else one fails to appreciate that the Cauchy-Riemann condition generalizes even to 3d real vector calculus and beyond, in terms of vector fields that are both divergenceless and curlless. $\endgroup$ – Muphrid Jun 1 '14 at 2:58
0
$\begingroup$

After poking and prodding, I found that if I define the derivative as being able to find two complex numbers $L$ and $M$ such that:

$$\lim_{h\to 0} \frac{f(z+h)-f(z)-Lh-M\overline{h}}{h}=0$$

then this is exactly equivalent to treating $f$ as a function from $\mathbb{R}^2$ to $\mathbb{R}^2$ and linear transformation $\mathbf{J}$ such that

$$\lim_{h\to 0} \frac{\| f(z+h)-f(z)-\mathbf{J}h\|}{\| h\|}=0$$

In this case, by analogy, it makes sense to write $L=\dfrac{\partial f}{\partial z}$ and $M=\dfrac{\partial f}{\partial \overline{z}}$, even when the function isn't complex differentiable in the usual sense.

This has to do with the basis matrices in the linked post: $$ \left\{\begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}0&1\\-1&0\end{bmatrix}\right\}$$ (linear combinations of which are linear transformations representing multiplication by a complex number)

and:

$$\left\{\begin{bmatrix}1&0\\0&-1\end{bmatrix},\begin{bmatrix}0&1\\1&0\end{bmatrix}\right\}$$ (linear combinations of which are linear transformations representing first complex conjugation then multiplication by a complex number)

The four of these are general enough so that linear combinations of them can represent any $L$ and $M$ acting on $h$ in the equation or to represent any 2x2 matrix $\mathbf{J}$.

The best part is that this notation allows one to write $f(z+\Delta z)\approx f(z)+\frac{\partial f}{\partial z} \Delta z+\frac{\partial f}{\partial \overline{z}} \overline{\Delta z}$

However, the fact that no complex number exists that allows one to write $\frac{df}{dz}\Delta z=\frac{\partial f}{\partial z} \Delta z+\frac{\partial f}{\partial \overline{z}} \overline{\Delta z}$ reaffirms that it's probably a bad idea to abuse this notation, and instead it's best to either stick to $\mathbb{R}^2\to \mathbb{R}^2$ and linear transformations, or functions for which $\frac{df}{dz}$ exists, and avoid switching between the two.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy