TITLE: Proof of Sufficiency of Cauchy-Riemann equations
QUESTION [5 upvotes]: I understand that the Cauchy-Riemann equations $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$ and $$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$ are necessary for a complex function to be complex-differentiable, but I would like to see a proof that they are also sufficient for this to be true. I am assuming that these partial derivatives exist and are continuous everywhere that they are satisfied.

REPLY [4 votes]: Well, let be provide some inspiration based on linear algebra. 
Remember in multivarible culculus, we always deem derivative as a linear mapping, or more concretely, a matrix. Now let's apply it in the case of complex linear space.
Let $f$ be a complex function on $ \mathbb{C} $, then $f$ is homolophic at point $z_0\in \mathbb{C}$ if and only if $Df(z_0)$ is a complex linear map from $\mathbb{C}$ to $\mathbb{C} $. Then identify $i$ to a matrix $\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}$, i.e., we deem $i$ as a complex linear map which rotate the plane 90 degrees anticlockwise. Next,  $Df(z_0)$ is a complex linear map if and only if $Df(z_0)$ commutes with $i$(think it why?), i.e., $Df(z_0)=\begin{pmatrix}
u_x & u_y \\
v_x & v_y
\end{pmatrix}$
$$\begin{pmatrix}
u_x & u_y \\
v_x & v_y
\end{pmatrix}\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}=\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}\begin{pmatrix}
u_x & u_y \\
v_x & v_y
\end{pmatrix}$$
if and only if $u_x=v_y, u_y=-v_x.$