Conside a complex function $ f(z)=u(x,y)+iv(x,y) $.
It is well known that if $ f $ is $ \mathbb{C} $ – differentiable, then the partial derivatives $ f_x, f_y $ exists, and satisfies Cauchy-Riemann equations.
Also, it is well known that if the partial derivatives exists, continuous at a point $ z $, and satisfies the Cauchy-Riemann equations, then the function $ z $ is $ \mathbb{C}$-differentiable at the point $ z $.
I noticed that if the partial derivatives exists and continuous at a point $z$, one can conclude that the function $ f:\mathbb{R}^2\ \to \mathbb{R}^2 $ is differentiable as a real function.
So, the question is : If a function $f(x,y): \mathbb{R}^2 \to \mathbb{R}^2 $ is differentiable as a real function, and the partial derivatives satisfies the Cauchy-Riemann equations, can we say that $ f $ is $\mathbb{C}$-differentiable as a complex function?
Thanks in advance.
Best Answer
Yes, you don't need the continuity of the partial derivatives.
Indeed, if $f$ is differentiable at a point $z_0$, then $$f(z_0+h)=f(z_0)+\mathrm{df}(z_0)(h) + o(h)$$
Denoting $z_0=x_0+iy_0$, and using Cauchy-Riemann equations, you get $$f(z_0+h)=f(z_0)+\frac{\partial f}{\partial x}(x_0,y_0) \times h + o(h)$$
so $f$ is $\mathbb{C}-$differentiable at $z_0$ (and $f'(z_0)=\frac{\partial f}{\partial x}(x_0,y_0) \times h = \mathrm{df}(z_0)(1)$)