Proof continuity of derivates using the Cauchy's integral

Question

I have been thinking about this and i have not reached any conclusion. If someone can tell me some tip to start.

I need to proof with the Cauchy's integral that if a function $f=u+iv$ that is holomorphic in a point $z_0$, the partial derivates $u_x,v_x,u_y,v_y$ there are all continuous at $z_0$.

Cauchy's integral formula. $$ f^{k}(z_0)=\frac{1}{2\pi i} \oint \frac{f(z) dz}{(z-z_0)^{k+1}}$$

Answer 1

Hint:

The integrand is a continuous function in $(z,z_0)$ on a compact neighborhood of the circle. So it is uniformly bounded. Apply dominated convergence theorem on it and you obtain that the integral is continuous in $z_0$.

Answer 2

Well, by definition since the function is holomorphic at the point $z_0$ then it is differentiable at a neighborhood of that point. and by Cauchy's integral formula we know that not only is it differentiable but it's arbitrary many times differentiable. on the other hand by Cauchy-Riemann equations we know that the first derivative equals $ u_x + iv_x$ and they guarantee for holomorphic functions that $u_x=v_y$ and $u_y=-v_x$ and since the derivative functions are continuous(because we know by Cauchy's integral formula that they are differentiable therefore continuous) both $u_x, v_x$ are continuous and by Cauchy-Reimann equations we get that $u_y , v_y$ are also continuous since they are equal to the former partial derivatives.

I hope I was able to be clear and help u.