What is the meaning of the derivative of a complex function.

Question

For the derivative of the complex function, when it is analytic, then it satisfies C-R equation. So in this case we have $f'(z_{0})=u_{x}(x_{0},y_{0})+iv_{x}(x_{0},y_{0})$. But if we consider the Vector valued function $g:\mathbb{R^{2}}\rightarrow\mathbb{R^{2}}$, the derivative at $z_{0}=(x_{0},y_{0})$ is $g'(z_{0})$, which is a Jacobi matrix. What makes these two things different? Is because the binary operation "vector multiplication" makes different sense in complex plane and $\mathbb{R^2}$?

Answer 1

When considered as a function from $\Bbb{R}^2$ to $\Bbb{R}^2$, a complex function differentiable at a point $z = x + iy$ must be differentiable at $(x, y)$, so in this sense, complex differentiability is stronger than $\Bbb{R}^2$-differentiability. Further, it's strictly stronger, as the Jacobean has to be of the form $$\begin{pmatrix} a & -b \\ b & a \end{pmatrix}.$$ In fact, this matrix corresponds to a complex derivative $a + ib$. The above matrix represents the matrix for the linear transformation of $z \mapsto (a + ib)z$, considered as maps from $\Bbb{C}$ to $\Bbb{C}$, with the ground field $\Bbb{R}$, over the basis $(1, i)$.

One consequence of this is that complex differentiable functions need to be conformal; the above matrix is a product of a scaling matrix $\sqrt{a^2 + b^2} I$ and a rotation matrix. This means that the map locally preserves angles, making it conformal. This is not true for general differentiable functions from $\Bbb{R}^2$ to $\Bbb{R}^2$!

Answer 2

If we let $\phi:\mathbb{C} \to \mathbb{R}^2$ be $\phi(x+iy) = (x,y)^T$ then $g(x) = \phi(f(x_1+ix_2))$.

Then ${\partial g(x) \over \partial x_1} = \lim_{t \to 0} \phi({f(x_1+i x_2 +t)- f(x_1+ix_2) \over t} ) = \phi(f'(x_1+i x_2))$ and ${\partial g(x) \over \partial x_2} = \lim_{t \to 0} \phi({f(x_1+i x_2 +it)- f(x_1+ix_2) \over t} ) = \phi(if'(x_1+i x_2))$.

In particular, if we write ${\partial g(x) \over \partial x_1} = \begin{bmatrix} a & b \end{bmatrix}$ then we must have ${\partial g(x) \over \partial x_2} = \begin{bmatrix} -b & a \end{bmatrix}$.