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

Question

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.

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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 $(x,y)=(0,0)$. ]

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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.

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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?

Answer 1

$\mathbb{C}$ 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.

Answer 2

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.

Answer 3

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 $f'(x_0+y_0i)=a+bi$.