Proof of Sufficiency of Cauchy-Riemann equations 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.
 A: 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 $u_x=v_y, u_y=-v_x.$
A: A function $f:\mathbb{R}^2 \to \mathbb{R}^2$ represents a complex-linear map (with respect to the standard complex-linear structure on $\mathbb{R}^2$ given by $i\cdot(u,v) = (-v,u)$) iff its matrix with respect to the standard $\mathbb{R}$-basis has the form $\begin{pmatrix}
a & -b \\
b & a
\end{pmatrix}$ (easy exercise). If $f(x,y) = (u(x,y),v(x,y))$ is $C^1$ then its derivative (again in the standard basis) is the matrix $\begin{pmatrix}
u_x & u_y \\
v_x & v_y
\end{pmatrix}.$
Now note that $f$ is complex-differentiable at a point if and only if it is real-differentiable there and its derivative is a complex-linear map. (Proof: $f$ is complex-differentiable at $z$ with $f'(z) = a \in \mathbb{C}$ if and only if $f(z+h) = f(z) + ah + o(h),$ and any $\mathbb{C}$-linear map $\mathbb{C} \to \mathbb{C}$ has the form $h \mapsto ah$ for some $a \in \mathbb{C}.$) Combining this with the remarks above we see that a function $f:\mathbb{R}^2 \to \mathbb{R}^2$ is complex-differentiable if and only if it is real-differentiable and its real and imaginary parts satisfy the C-R equations. In particular if $f$ is of class $C^1$ then $f$ is complex-differentiable if and only if its real and imaginary parts satisfy the C-R equations.
A: The integral around every single triangle is zero by Green's theorem. If the integral around every single triangle is zero then the function is complex differentiable by Morera's theorem. Done.
