Geometrical interpretation of complex differentiability I want to know  the geometrical view of  difference between differentiability of two variable functions in $\mathbb R^2$ and  the meaning of differentiability for holomorphic functions,I know the solution with algebra and analysis tools but I don't know what will happen for their geometry
 A: Both real and complex differentiability boil down to the same concept: a function $f:\mathbb C\to \mathbb C$ is differentiable in $z_0$ if there exists a linear map $L:\mathbb C\to\mathbb C$ which is a good approximation of $f$ at $z_0$. More specifically, it means that
$$\lim_{h\to0}\frac{f(z_0+h)-f(z_0)-L(h)}{\vert h\vert}=0.$$
$f(z_0)+L(h)$ is the linear approximation of $f$ at $z_0$, and the above equation says that the difference between $f$ and its linear approximation is in some way small when we are close to $z_0$ (meaning that $h$ is small). The difference between the the types of differentiability is subtle, but important. It's in the definition of "linear". There are two ways to view $\mathbb C$ as a vector space. Either as a two-dimensional vector space over $\mathbb R$, or a one-dimensional vector space over $\mathbb C$. And the two notions will result in different kinds of linearity, called $\mathbb R$-linearity and $\mathbb C$-linearity. $\mathbb R$-linear maps are the ones you know well from linear algebra: Rotations, dilations, shearings, and more. But $\mathbb C$-linear maps are more restricted. A map $L:\mathbb C\to\mathbb C$ is $\mathbb C$-linear, if for all $\lambda, z,w\in\mathbb  C$ it holds that $L(z+w)=L(z)+L(w)$ and $L(\lambda z)=\lambda L(z)$. Basically like the usual linear maps, but the scalars we can pull out of $L$ are now complex instead of just real. And this is big, because this now means that $L(z)=L(1\cdot z)=L(1)\cdot z$. And we can choose $L(1)$ to our liking, so the $\mathbb C$-linear maps are just the multiplications by a complex number. And those are known to be rotation-dilations, geometrically: Multiplying a complex number $z$ by another complex number $c$ rotates and dilates $z$ by an amount determined by $c$ (by the way, these are also $\mathbb R$-linear, but not all $\mathbb R$-linear maps are also $\mathbb C$-linear). And that's the big geometric difference between real and complex differentiability: A function is real differentiable if it can be approximated by any linear map. And it is complex differentiable if it can be approximated by a rotation-dilation, which is a very specific kind of linear map. Also, rotation-dilations preserve angles between smooth curves, unless they are the zero-map. And holomorphic functions do so, too: Take any two regular, smooth curves. If we apply a holomorphic map whose derivative is nowhere $0$ to the complex plane, then we get two different curves. But if they intersected beforehand, then they will still intersect, and the angle of intersection will be the same.
A: (See figure below)
Consider the jacobian matrix of a function $(u(x,y),v(x,y))$, $\mathbb{R}^2\to \mathbb{R}^2$:
$$J=\begin{pmatrix}\partial u/\partial x&\partial u/\partial y\\ 
 \partial v/\partial x&\partial v/\partial y\end{pmatrix}\tag{0}$$
In the case of a complex differentiable function, its entries must obey the Cauchy-Riemann equations,
$$\partial u/\partial x=\partial v/\partial y \  \ \ \ \text{and} \ \ \ \ \partial u/\partial y=-\partial v/\partial x$$
It means that $J$, having the structure:
$$J=\begin{pmatrix}a&-b\\b& \ \ a\end{pmatrix}\tag{1}$$
can be interpreted as the matrix of a similitude ( = rotation followed by homothety). Indeed, setting $a=r \cos \theta, b=r \sin \theta$, we obtain:
$$J=r\begin{pmatrix}\cos \theta&-\sin \theta\\ \sin \theta& \ \ \cos \theta\end{pmatrix}\tag{2}$$
What is the geometrical meaning of these parameters $r$ and $\theta$ ? Let us take the example of function $$Z=f(z)=\sqrt{z}\tag{2}$$
Horizontal and vertical lines are mapped by transformation (2) onto branches of hyperbolas. For example, the (blue) branch of hyperbola closest to the origin is the image of the (blue) vertical line $x=1$ ; in particular, point $(1,0)$ is its own image... The orthogonality of curves is preserved (a well known property of the so-called conformal mappings).
I have singled out point $z_0=3+4i$ which is at the intersection of blue line $x=3$ and red line $y=4$. We find its image $\sqrt{z_0}$ at the intersection of the $3$rd blue hyperbola and the 4th red hyperbola. Of course, its true value is by reference to coordinate axes $(u,v)$, and we read $\sqrt{z_0}=2+i.$
let us now, connect this with what has been said before: consider $2$ small vicinities around $z_0$ and its image  $\sqrt{z_0}$, materialized by green squares; the second green square is the image of the first one by a rotation and a homothety. The scale in the right figure being about 4 times bigger than in the left figure, the homothety ratio $r$ should be around $0.25$. We can also anticipate on the rotation angle which should be around $-30$ degrees (minus sign is important!).
Let us do now precise computations that will confirm that the approximate values we have obtained above were in the good range.
As the derivative of $f(z)=\sqrt{z}$ is $f(z)=1/(2\sqrt{z})$ (same formula than for its real cousin), this gives for $z=z_0$:
$$f'(z_0)=1/(2 \sqrt{z_0})=1/(2(2+i))=0.2-0.1i=re^{i\theta}$$
whose module $r=0.2236$ account for the homothety ratio (shrinking here), and argument
$$\theta \ = \ -\arctan \frac12 \ = \ -0.4636 \ \text{radians} \ = \ -26.56 \ \text{degrees}$$
Our guesses weren't bad!
Remark 1: All this has been made possible because a matrix such as (1) or (2) "is" a complex number resp. under its algebraic and trigonometric form, but a complex number under its "dynamic aspect" acting (by multiplication) on its static colleagues...
Remark 2: One could object that a translation should also have been taken into account, but this is not the case: for differentiability, we consider that all is done at the origin as well in the original plane and in the image plane.

Fig.1 : "Conformal mapping" associated with function $f(z)=\sqrt{z}$. The images of horizontal and vertical lines $x=x_0$ and $y=y_0$ are hyperbolas with equations $u^2-v^2=x_0$ and $2uv=y_0$ resp. (with corresponding colors).
A: I try to explain the meaning of $f'(z)dz$ for complex differentiable functions at a point. Let the complex numbers be interpreted in the plane as arrows starting from the origin and making some angle with $x$-axis. Their addition is clear (I hope) but their multiplication has an arrow with a lenght as the product of the lenghts and an angle as the sum of the angles. Let us write $dz = re^{i\varphi}$ and $f'(z) = Re^{i\Phi}$. Their product is
$$
f'(z)dz = re^{i\varphi} Re^{i\Phi} = rR e^{i(\varphi + \Phi)}
$$
This product should represent the small change in $f(z)$ as response to a sufficiently small change $dz$ in $z$. That is
$$
\Delta f(z) \approx df(z) = rR e^{i(\varphi + \Phi)}
$$
Now hold $r,R$ and $\Phi$ fixed and change $\varphi$ you see that the change in $f(z)$ (or more precisely $df(z)$) rotates  with the same angle $\varphi$. That is $df(z)$ and $dz$ retain the angle between them. Allow now only $r$ to change, then $df(z)$ and $dz$ are scaled by the same number.
This dependence of $df(z)$ on $dz$ corresponds to a linear transformation namely a rotation + scaling by the angle of $dz$ and by its length respectively. Now the derivative in $\mathbb R^n$ is a linear transformation. The one that we have in $\mathbb C$ is a special case of that in $\mathbb R^2$. For more on linear transformations in $\mathbb R^2$ see this and this an remember that complex derivative is a special one.
A: If $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is smooth, then by definition it has a Jacobian J.  For $f$ to be complex-differentiable at a point $(x_0,y_0)$, there must exist a complex number $w \in \mathbb{C}$ equivalent to $J$ in the sense that for all $x,y \in \mathbb{C}$, $$J\begin{bmatrix}
           x \\
y
\end{bmatrix} = w(x + iy)$$ where multiplication on the left is the usual matrix multiplication  but multiplication on the right is multiplication in $\mathbb{C}$.  The linear transformations $J$ equivalent to multiplying by some $w$ are very special; they are formed by a rotation and scaling. Tristan Needham, in his book "Visual Complex Analysis" calls these transformations (made up of rotation + scaling) amplitwists.
An informal way of saying this: is that $f$ is real-differentiable at $z$ if it has the property that when you look at $f$ at a very small neighborhood around $z$ then $f$ starts to look like it is linear; while $f$ is complex-differentiable at $z$ if it has the property that when you look at $f$ at a very small neighborhood around $z$ then $f$ starts to not only linear, but like an "amplitwist". Under a magnifying glass, $f$ is just rotating and scaling.
