Let $f=u+i v$ be a complex function. Then the real part $u=u(x,y)$ is a multivariable real-valued function. It is well known that if $f$ is holomorphic at $z=x+iy$, then $u$ is $C^1$ near $(x,y) \in \mathbb{R}^2$.
I was wondering if $u$ is differentiable at $(x,y)$ when $f$ is complex-differentiable at a point $z=x+iy.$ Would you give me any comment of it. Thanks in advance!
On
Yes , by Caucht-Rieman , $f$ is complex-differentiable at $z_0= (x_0,y_0)$ iff $u,v$ both differentiable at $(x_0,y_0)$ and satisfy the Caucht-Rieman equations
$\dfrac{\partial u}{ \partial x} =\dfrac{\partial v}{ \partial y} ,\dfrac{\partial u}{ \partial y} = -\dfrac{\partial v}{ \partial x} $
Let $z_0=x_0+iy_0$ and suppose that $f$ is complex differentiable at $z_0$. Let $h=h_1+ih_2 \ne 0$ and $f'(z_0)=a+ib$.
We then have that
$Q(h):= \frac{f(z_0+h)-f(z_0)-hf'(z_0)}{|h|} \to 0$ as $h \to 0$.
An easy computation shows that
$R(h):=Re (Q(h))$ is given by
$R(h)=\frac{u(x_0+h_1,y_0+h_2)-u(x_0,y_0)-(h_1a-h_2b)}{|h|}$.
Since $Q(h) \to 0$ as $h \to 0$, we get
$R(h) \to 0$ as $h \to 0$, which shows that $u$ is real differentiable at $(x_0,y_0)$ with $u_x(x_0,y_0)=a$ and $u_y(x_0,y_0)=-b$.