Derivative of a Complex Function as a $\mathbb{C}$-linear Transformation I'm trying to show that the derivative of a differentiable complex function is a $\mathbb{C}$-linear mapping of $\mathbb{C}$ to itself, and since every $\mathbb{C}$-linear map is of the form
$$ 
\begin{bmatrix}
c_1 & c_2 \\
-c_2 & c_1
\end{bmatrix}
$$
then I can deduce Cauchy-Riemann equations.
In particular I think when $f$ is differentiable in a neighborhood of $a$, it's enough to show that $Df (a)$ respects scalar multiplication over $\mathbb{C}$ but my efforts always need Cauchy-Riemann equations to prove the desired which means it doesn't work since we want to show them.
In addition I mean a complex-valued function differentiable at neighborhood of $a$, if the limit $\lim_{h \rightarrow0}\frac{f(x+h)-f(x)}{h}$ exists for every $x$ in that neighborhood.
 A: Differentiability is a pointwise issue, you we only need to worry about what is happening at $a$. Suppose $f:U \to \Bbb{C}$ is a  function defined on an open subset of the complex plane, $a\in U$, and that $f'(a):= \lim\limits_{h\to 0}\frac{f(a+h)-f(a)}{h}$ exists. I hope you can convince yourself that this is equivalent to
\begin{align}
\lim_{h\to 0}\dfrac{f(a+h) - f(a) - f'(a)\cdot h}{h} = 0.
\end{align}
Or equivalently, we can put absolute values everywhere
\begin{align}
\lim_{h\to 0}\dfrac{\left|f(a+h) - f(a) - f'(a)\cdot h\right|}{|h|} = 0.
\end{align}
Look at what we have inside; the function $h\mapsto f'(a)\cdot h$ is a $\Bbb{C}$-linear transformation $\Bbb{C}\to \Bbb{C}$. This means exactly that $Df_a(h)= f'(a)\cdot h$, and hence that $Df_a:\Bbb{C}\to \Bbb{C}$ is $\Bbb{C}$-linear.
Finally, since every $\Bbb{C}$-linear transformation has that matrix representation you describe (relative to the  basis $\{1, i\}$ of the $2$-dimensional vector space $\Bbb{C}$ over the field $\Bbb{R}$), you can deduce the Cauchy-Riemann equations.
