Functions of a complex variable and the Cauchy-Riemann equations If I have a complex-valued function of a complex variable $f(s)$, where $s\in\mathbb{C}$, which does not satisfy the Cauchy-Riemann equations, then does it make sense to actually take the derivative of the function?
On taking the derivative would it make sense to then use the resulting equations to reach logical conclusions, whatever they might be? Or would any such conclusions be invalid since the Cauchy-Riemann equations are not satisfied?
 A: It depends. Your function $f : \mathbb{C} \rightarrow \mathbb{C}$ can always be seen as a function from $\mathbb{R}^2$ to $\mathbb{R}^2$. If it is at least differentiable in the real sense, then it certainly makes sense to compute the derivative of $f$ as a real function, which will give you a $2x2$ real matrix.
The relation this derivative has with the complex derivative can be expressed in many ways, such as the following: you can define the operators
$$\frac{\partial}{\partial{}z} = \frac{1}{2}\left(\frac{\partial}{\partial{x}} -i\frac{\partial}{\partial{y}}\right),~~\frac{\partial}{\partial{}\bar{z}} = \frac{1}{2}\left(\frac{\partial}{\partial{x}} +i\frac{\partial}{\partial{y}}\right),$$
which can be applied to any function $f$ having at least partial derivatives. Then, you can check that such a function $f$ satisfies the Cauchy-Riemann equations if and only if
$$\frac{\partial{f}}{\partial{\bar{z}}} = \frac{1}{2}\left(\frac{\partial{f}}{\partial{x}} +i\frac{\partial{f}}{\partial{y}}\right) = 0.$$
You can also check that the matrix of $\mathrm{df}$ as a map $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a conformal linear mapping at every point if and only if $f$ satisfies Cauchy-Riemann. A conformal linear mapping is a linear mapping that preserves angles, intuitively speaking. Formally, $T : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is conformal if and only if it has the form $\alpha{}L$, where $L : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is an orthogonal linear map (that is, $L^tL = 1$).
