# Difference of $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ and as $f: \mathbb{C} \rightarrow \mathbb{C}$.

I want to know the difference of differentation as $$f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$$ and $$f: \mathbb{C} \rightarrow \mathbb{C}$$.

What are their differences, $$f$$ as two real variables, or $$f$$ as differentiation as a complex function?

This question arose when I took the youtube lectures by "Richard E. BORCHERDS" on complex analysis.

First treatment of real analysis :

In multivariable calculus, when we set $$f(x,y)$$ its total derivatives is written as \begin{align} df =f_x dx + f_y dy \end{align} where $$f_x, f_y$$ are partial derivatives with respect to $$x,y$$

Formally, we say that a function $$f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$$ is differentiable at $$a \in \mathbb{R}^2$$ if it exists a continuous linear map $$\nabla f(a) : \mathbb{R}^2 \rightarrow \mathbb{R}^2$$ such that \begin{align} \lim_{h \rightarrow 0} \frac{f(a+h) - f(a) - \nabla f(a) \cdot h}{\|h\|} =0 \end{align} so when we consider multivariable calculus, we have to see whether the multivariable function have a partial derivatives(or directional derivatives) and then see the above limit holds[In the calculus, we learn that a function having a partial derivatives but not differentiable, i.e., $$f(x,y) = \frac{xy}{\sqrt{x^2+y^2}}$$ at $$(x,y) \neq (0,0)$$ but $$0$$ at $$(x,y)=(0,0)$$. ]

In the complex analysis, we treat $$f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$$ or $$f: \mathbb{C} \rightarrow \mathbb{C}$$ and define complex derivatives analogus to real derivatives and obtain Cauchy Riemann equation.

For example $$w=u+iv$$, \begin{align} \begin{pmatrix} u(x,y) \\ v(x,y) \end{pmatrix} = \begin{pmatrix} u(x_0, y_0) \\ v(x_0, y_0) \end{pmatrix} + \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix} \begin{pmatrix} x-x_0 \\ y-y_0 \end{pmatrix} + \epsilon \end{align} and doing $$w$$ as \begin{align} w=w_0 + A (z-z_0) + \epsilon, \quad A \in \mathbb{C} \end{align} [This is Borcherds treatment of differentiation as a linear approximation. Like real case he treats $$w$$ as $$\mathbb{C}$$ and does the linear approximation on $$\mathbb{C}$$] then identifying the component of $$A$$ he obtain Cauchy Riemann equation.

In complex cases, I feel Borcherds treat the differentiation as $$x,y$$ and $$z$$ equally, but in general case those two approaches are different am I?

For example, when dealing with complex analysis, differentiable at some open region (analytic) implies $$C^{\infty}$$ but I know in multi-variable calculus this does may not happen.

What are their differences, $$f$$ as two real variables, or $$f$$ as differentiation as a complex function?

$$\mathbb{C}$$ is literally $$\mathbb{R}^2$$ with additional vector multiplication. The complex $$i$$ is simply $$(0,1)$$. For $$a,b\in\mathbb{R}$$ we can easily then check that $$a+bi$$ is the same as $$(a,b)\in\mathbb{C}$$. And so a function $$f:\mathbb{C}\to\mathbb{C}$$ is literally the same as a function $$f:\mathbb{R}^2\to\mathbb{R}^2$$.

But complex and real differentiation is (somewhat) different. For starters their respective definitions are obviously different. But every complex differentiable function $$f:\mathbb{C}\to\mathbb{C}$$ is real differentiable. Moreover if $$f(a+bi)=u(a+bi)+iv(a+bi)$$, where $$u,v:\mathbb{C}\to\mathbb{R}$$ are real valued functions and $$f$$ is complex differentiable, then

$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$ $$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$

which are known as Cauchy-Riemann equations. It turns out that this is also a sufficient condition for $$f$$ to be complex differentiable, given that both $$u,v$$ are continuously real differentiable. In such situation both derivatives agree in the following sense: since $$\nabla f(a):\mathbb{R}^2\to\mathbb{R}^2$$ is a linear map, then it corresponds to a real $$2\times 2$$ matrix, which then corresponds to a complex number since in this situation our matrix has a specific form $$\left[\begin{matrix}\alpha & \beta \\ -\beta & \alpha\end{matrix}\right]$$. The $$\alpha+\beta i$$ complex number is our complex derivative at $$a$$ and vice versa.

And so you can think of complex differentiation as a very special case of real differentiation. In fact those two simple equations make the complex analysis much much more restrictive than its real counterpart.

For example as you said: for complex differentiation $$C^1$$ already implies $$C^\infty$$ (smooth) and even $$C^\omega$$ (analytic). Another difference is that every bounded complex differentiable function must be constant (Liouville's theorem). Even more: a complex differentiable function takes every possible complex value, except at most one, if non-constant (little Picard theorem), and so on.

The derivative of a function $$\mathbb R^2 \to \mathbb R^2$$ is a $$2\times2$$ matrix. The complex derivative of a function $$\mathbb C \to \mathbb C$$ is a complex number. Application of the $$2\times2$$ matrix derivative is analogous to multiplication by the complex derivative.

If $$a + bi$$ is a complex number, then multiplying the complex number $$x + yi$$ by $$a + bi$$ sends it to $$(ax -yb) + (ay + bx)i$$. That means that the complex number can itself be thought of as the following $$2\times2$$ matrix.

$$\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$$

This leads us to the following definition of complex differentiability. A function $$f \colon \mathbb C \to \mathbb C$$ is complex differentiable at a point $$z$$ if it is differentiable at $$z$$ when considered as a function $$\mathbb R^2 \to \mathbb R^2$$ and if its its derivative at $$z$$ is a matrix taking the form $$\begin{pmatrix} a & -b \\ b & a \end{pmatrix}\,.$$ Therefore: any complex differentiable function is differentiable in the real sense, but not every function that is differentiable in the real sense is complex differentiable, since not every $$2\times 2$$ matrix takes the form given above.

Now, since the coefficients of the derivative of a function $$f = \langle u,v\rangle$$ are precisely the partial derivatives $$\begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix}$$ then this statement is precisely the Cauchy-Riemann equations.

If $$f\colon\Bbb C\longrightarrow\Bbb C$$ is differentiable, then it is also differentiable as a map from $$\Bbb R^2$$ into $$\Bbb R^2$$. Besides, if $$f'(z_0)=c$$, then, if you see $$f$$ as a map from $$\Bbb R^2$$ into $$\Bbb R^2$$, if $$z_0=x_0+y_0i$$, and if $$c=a+bi$$ (with $$x_0,y_0,a,b\in\Bbb R$$), then $$f'(x_0,y_0)$$ is the linear map whose matrix with respect to the standard basis is$$\begin{bmatrix}a&-b\\b&a\end{bmatrix}.\tag1$$That's why, in general, if $$f\colon\Bbb R^2\longrightarrow\Bbb R^2$$ is differentiable, then it is not differentiable as a map from $$\Bbb C$$ into $$\Bbb C$$; in general, the Jacobian of $$f$$ at a point $$(x_0,y_0)$$ is not of the form $$(1)$$. But if it is (and if $$f$$ is a $$C^1$$ function), then $$f$$ will actually be differentiable at $$x_0+y_0i$$, and $$f'(x_0+y_0i)=a+bi$$.