Understanding proof of theorem 2.4 in Stein Complex analysis 
Suppose $f = u+iv$ is a complex-valued function defined on an open set $\Omega$. If $u$ and $v$ are continuously differentiable and satisfy the Cauchy-Riemann equations on $\Omega$ then $f$ is holomorphic on $\Omega$ and $f'(z) = \partial f/\partial z$.


In the proof, they first write $u(x+h_1,y+h_2)-u(x,y) =\frac{\partial u}{\partial x}h_1+\frac{\partial u}{\partial y}h_2+|h|\psi_1(h)$. How can I just write like that?
 A: You can write $u(x+h_1,y+h_2)$ as $$u(x,y)+\bigl(u(x+h_1,y+h_2)-u(x,y+h_2)\bigr)+\bigl(u(x,y+h_2)-u(x,y)\bigr).$$But, by the mean value theorem, there are maps $\theta_1,\theta_2\colon\Omega\longrightarrow[0,1]$ such that\begin{multline}u(x+h_1,y+h_2)-u(x,y+h_2)=h_1\frac{\partial u}{\partial x}\bigl(x+\theta_1(x,y)h_1,y+h_2\bigr)=\\=h_1\frac{\partial u}{\partial x}(x,y)+h_1\left(\frac{\partial u}{\partial x}\bigl(x+\theta_1(x,y)h_1,y+h_2\bigr)-\frac{\partial u}{\partial x}(x,y)\right)\end{multline}and that\begin{align}u(x,y+h_2)-u(x,y)&=h_2\frac{\partial u}{\partial y}(x,y+\theta_2(x,y)h_2)\\&=h_2\frac{\partial u}{\partial y}(x,y)+h_2\left(\frac{\partial u}{\partial y}(x,y+\theta_2(x,y)h_2)-\frac{\partial u}{\partial y}(x,y)\right).\end{align}Since $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ are continuous, if you define $\varphi_1,\varphi_2\colon\Omega\longrightarrow\Bbb R$ by$$\varphi_1(h_1,h_2)=\frac{\partial u}{\partial x}\bigl(x+\theta_1(x,y)h_1,y+h_2\bigr)-\frac{\partial u}{\partial x}(x,y)$$and$$\varphi_2(h_1,h_2)=\frac{\partial u}{\partial y}(x,y+\theta_2(x,y)h_2)-\frac{\partial u}{\partial y}(x,y),$$you have $\varphi_1(0,0)=\varphi_2(0,0)=0$, $\varphi_1$ and $\varphi_2$ are continuous at $(0,0)$, and$$u(x+h_1,y+h_2)=u(x,y)+h_1\frac{\partial u}{\partial x}(x,y)+h_2\frac{\partial u}{\partial h}(x,y)+h_1\varphi_1(h_1,h_2)+h_2\varphi_2(h_1,h_2).$$So, define $\psi(h_1,h_2)=\sqrt{\varphi_1^{\,2}(h_1,h_2)+\varphi_2^{\,2}(h_1,h_2)}$, and you're done.
