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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.

Masoud
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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.

peek-a-boo
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  • So here $Df(a) $ is the linear approximation of $f$ at $a$ and $f'(a) $ is the value of that limit? – Masoud Oct 29 '20 at 06:13
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    @Masoud yes, that's right. Also, note that in general whenever the domain of $f$ is an open subset of $\Bbb{R}$ or $\Bbb{C}$ (the target space can be any normed vector space), we always have that the linear approximation $Df_a$ exists if and only if $f'(a)$, the usual limit of the difference quotient, exists in which case $Df_a(h)=f'(a)\cdot h$. – peek-a-boo Oct 29 '20 at 10:47
  • What's the explicit definition of $Df(a) $ here? I always thought that for a function $f : \mathbb{R}^n \to \mathbb{R}^m$, $Df(a) $ and $f'(a) $ are same thing with different notations. But it seems that in complex plane these two are different essentially, but Cauchy-Riemann make them same things like general case. Am I right? – Masoud Oct 29 '20 at 16:55
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    @Masoud I don't like to confuse $Df(a)$ and $f'(a)$. Here, for any function $f:V\to W$ between Banach spaces, $Df_a$ is the linear approximation of $f$ at $a$, i.e it is the unique linear map $T:V\to W$ (if it exists) such that $\frac{\lVert f(a+h) - f(a) - T(h) \rVert_W}{\lVert h\rVert_V} \to 0$ as $h\to 0$. This is called the Frechet-derivative of $f$ at $a$, and it doesn't matter if the Banach spaces $V,W$ are both real or both complex, or finite or infinite-dimensional. If we assume $V=\Bbb{R}^n, W=\Bbb{R}^m$ then $Df_a:V\to W$ is a linear map between finite-dimensional cartesian spaces – peek-a-boo Oct 29 '20 at 17:46
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    As such, it has a matrix representation relative to the standard bases of $\Bbb{R}^n, \Bbb{R}^m$. This matrix representation is usually denoted $f'(a)$ (or as $J_f(a)$, or even $Df(a)$). Because I do not like to confuse a linear transformation with its matrix representation (even in cartesian spaces), I like to use separate notations $f'(a)$ vs $Df_a$. Some people like to use $f'(a)$ and $Df(a)$ interchangeably and think of either as linear transformations or matrices, but I do not like that for some reason. – peek-a-boo Oct 29 '20 at 17:50
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    So, simply by linear algebra, $Df_a(h)= f'(a)\cdot h$, where on the RHS we interpret it as multiplication of the matrix $f'(a)$ on the column vector $h$, and on the LHS, we have the evaluation of the linear map $Df_a$ on the vector $h$. So, really there is nothing special about the complex plane in this regard; it's just a matter of keeping the linear algebra straight in your mind (and once you know what exactly you're doing, you're free to use/abuse these notations as much as you wish). – peek-a-boo Oct 29 '20 at 17:52