Confusion in the identification of $\mathbb C$ with $\mathbb R^2$ in complex differentiability

Question

I'm reading about complex differentiation. The authors first identify $\mathbb C$ with $\mathbb R^2$, and $f \in \mathbb C^X$ with $F = (u,v) \in (\mathbb R^2)^X$.

Then they present a theorem and its proof:

1. The identification of $$\frac{\left|F\left(x_{0}+\xi, y_{0}+\eta\right)-F\left(x_{0}, y_{0}\right)-A(\xi, \eta)\right|}{|(\xi, \eta)|}$$ with $$\left |\frac{f\left(z_{0}+h\right)-f\left(z_{0}\right)-f^{\prime}\left(z_{0}\right) h}{h} \right |$$ makes sense to me.

However, I could not understand how their values when taking the limits are equal. We say that we identify $F\left(x_{0}+\xi, y_{0}+\eta\right)$ with $f(z_0+h)$, but we do not say that their values are equal. The value of the first expression is in $\mathbb R^2$, whereas the value of the second expression is in $\mathbb C$.

2. We have $z_0 = x_0 + i y_0$ and

$$[\partial f (z_0)]=\left[\begin{array}{cc}{\alpha} & {-\beta} \\ {\beta} & {\alpha}\end{array}\right]$$ and $$\left[\partial F\left(x_{0}, y_{0}\right)\right]=\left[\begin{array}{cc}{\partial_{1} u\left(x_{0}, y_{0}\right)} & {\partial_{2} u\left(x_{0}, y_{0}\right)} \\ {\partial_{1} v\left(x_{0}, y_{0}\right)} & {\partial_{2} v\left(x_{0}, y_{0}\right)}\end{array}\right]$$

I could not understand how we go from the identification of $\mathbb C$ and $\mathbb R^2$ to $[\partial f (z_0)]=\left[\partial F\left(x_{0}, y_{0}\right)\right]$.

Could you please elaborate on these points? Thank you for your help!

Answer 1

The value of $F(x)$ and $f(x)$ might not be equal since as you point out they are only identified in some way, but once we take norms, $|F(x)|=|f(x)|$ is an equality of real numbers. This is because the identification between $\mathbb{C}$ and $\mathbb{R}^2$ that we have specified preserves the norm.

Edit: The fact that $A$ satisfies the above limit means that $[\partial F(x_0,y_0)]=A$. This is simply the definition of the total derivative. If we look at what $A$ we see that it precisely the matrix of partials of $u$ and $v$ that you wrote. The way we get this limit is from the equality of norms I mentioned above.

Answer 2

1. Those are not the expressions for the derivatives and they are not being identified. It might be better to note that each is the difference of the difference quotient and the derivative it approaches if the function is actually differentiable at the point $(x_0, y_0)$, resp. $z$: $$ \frac{\left|F\left(x_{0}+\xi, y_{0}+\eta\right)-F\left(x_{0}, y_{0}\right)-A(\xi, \eta)\right|}{|(\xi, \eta)|} \\ = \left| \frac{F\left(x_{0}+\xi, y_{0}+\eta\right)-F\left(x_{0}, y_{0}\right)}{|(\xi, \eta)|} - \frac{A \cdot (\xi, \eta)}{|(\xi, \eta)|} \right| $$ and $$\left |\frac{f\left(z_{0}+h\right)-f\left(z_{0}\right)-f^{\prime}\left(z_{0}\right) h}{h} \right | \\ = \left| \frac{f\left(z_{0}+h\right)-f\left(z_{0}\right)}{h} - f^{\prime}\left(z_{0}\right) \right| $$ Setting these equal to zero is forcing that each of the limits of the difference quotient is the derivative -- i.e., the two derivatives actually exist. The information content in the equation is $... = 0$, that both of these things are going to zero.

2. Equation 2.7 shows how we are to associate a linear operator acting on $\Bbb{C}$ (i.e., multiplication by a constant) and the associated linear operator acting on $\Bbb{R}^2$ (multiplication by a matrix of a special structure).

   The derivative is a linear operator. Consequently, we identify the action of differentiation on a $\Bbb{C}$-function, $f$, with the action of a certain matrix acting on $F$, the $\Bbb{R}^2$ function identified with $f$. This is where we constrain the set of $\Bbb{R}^2$ functions to actually agree with $\Bbb{C}$ functions -- we only allow an $\Bbb{R}^2$ function, $F = (u,v)$, whose derivative, $\partial F = \begin{bmatrix} u_x & u_y \\ v_x & v_y \end{bmatrix}$, does the same thing (to a vector representing a complex number) as multiplying (that complex number) by a complex number using the identification in equation 2.7. I.e., we force the equality of the diagonal components and the antisymmetry of the off-diagonal components to specifically pick out the functions that correctly implement complex multiplication through the identification in equation 2.7.

Answer 3

From @user293794's answer, I seem to get something, but I'm not sure if my understanding is correct or not. I posted it here. It would be great if someone helps me verify it. Thank you so much!

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My understanding of the proof:

First, the authors identify $\mathbb C$ with $\mathbb R^2$ through canonical identification $$ \mathcal I:\mathbb{C} \to \mathbb{R}^{2}, \quad x+iy \mapsto ( x,y ) $$

It can be check that $\mathcal I$ is a topological isomorphism. For $f: X \to \mathbb C$, the authors define associated map $F:\mathcal I(X) \to \mathbb R^2$ by $F(x,y) = ( \operatorname{Re} f(z), \operatorname{Im} f(z) )$ for all $z=x+iy \in X$.

$\Longrightarrow$: Suppose $f$ is complex differentiable at $z_0 = x_0 + i y_0$. Let $\alpha=\operatorname{Re} f^{\prime}\left(z_{0}\right)$, $\beta=\operatorname{Im} f^{\prime}\left(z_{0}\right)$, and $$A=\left[\begin{array}{cc}{\alpha} & {-\beta} \\ {\beta} & {\alpha}\end{array}\right]$$

Let $h = \xi +i\eta \in \mathbb C$. It follows from $f$ is differentiable at $z_0$ that $$\lim _{h \rightarrow 0_{\mathbb C}} \frac{f\left(z_{0}+h\right)-f\left(z_{0}\right)-f^{\prime}\left(z_{0}\right) h}{\|h\|} = 0_{\mathbb C}$$

We have the following identifications:

• $0_{\mathbb C} \leftrightarrow 0_{\mathbb R^2}$

• $h \leftrightarrow (\xi ,\eta)$

• $f(z_0+h) \leftrightarrow \mathcal I (f(z_0+h)) = F(x_0+\xi, y_0+\eta)$

• $f(z_0) \leftrightarrow \mathcal I (f(z_0)) = F(x_0, y_0)$

• $f'(z_0)h \leftrightarrow A \cdot \begin{bmatrix} \xi \\ \eta \\ \end{bmatrix}$

Because $\mathcal I$ preserves the topological structure, we have $$\lim_{(\xi ,\eta) \to 0_{\mathbb R^2}} \frac{F(x_0+\xi, y_0+\eta) - F(x_0, y_0) - A \cdot \begin{bmatrix} \xi \\ \eta \\ \end{bmatrix}}{\|(\xi ,\eta)\|} = 0_{\mathbb R^2}$$

By the definition of total derivative, we get $\partial F(x_0,y_0) (\xi ,\eta)= A \cdot \begin{bmatrix} \xi \\ \eta \\ \end{bmatrix}$ for all $(\xi ,\eta) \in \mathbb R^2$. Hence the matrix representation of $\partial F(x_0,y_0)$ is $[\partial F(x_0,y_0)] = A$. On the other hand, we have $$\left[\partial F\left(x_{0}, y_{0}\right)\right] = \left[\begin{array}{cc}{\partial_{1} u\left(x_{0}, y_{0}\right)} & {\partial_{2} u\left(x_{0}, y_{0}\right)} \\ {\partial_{1} v\left(x_{0}, y_{0}\right)} & {\partial_{2} v\left(x_{0}, y_{0}\right)}\end{array}\right]$$

As such, $$A = \left[\begin{array}{cc}{\partial_{1} u\left(x_{0}, y_{0}\right)} & {\partial_{2} u\left(x_{0}, y_{0}\right)} \\ {\partial_{1} v\left(x_{0}, y_{0}\right)} & {\partial_{2} v\left(x_{0}, y_{0}\right)}\end{array}\right]$$

$\Longleftarrow$: Suppose $F$ is differentiable at $(x_0,y_0)$ and satisfies Cauchy-Riemann equations, i.e. $\partial_{1} u\left(x_{0}, y_{0}\right)=\partial_{2} v\left(x_{0}, y_{0}\right)$ and $\partial_{2} u\left(x_{0}, y_{0}\right)=-\partial_{1} v\left(x_{0}, y_{0}\right)$.

Let $a = \partial_{1} u\left(x_{0}, y_{0}\right)+i \partial_{1} v\left(x_{0}, y_{0}\right) \in \mathbb C$ and $(\xi ,\eta) \in \mathbb R^2$.

It follows from $F$ is differentiable at $(x_0,y_0)$ that $$\lim_{(\xi ,\eta) \to 0_{\mathbb R^2}} \frac{F(x_0+\xi, y_0+\eta) - F(x_0, y_0) - \left[\begin{array}{cc}{\partial_{1} u\left(x_{0}, y_{0}\right)} & {\partial_{2} u\left(x_{0}, y_{0}\right)} \\ {\partial_{1} v\left(x_{0}, y_{0}\right)} & {\partial_{2} v\left(x_{0}, y_{0}\right)}\end{array}\right] \cdot \begin{bmatrix} \xi \\ \eta \\ \end{bmatrix}}{\|(\xi ,\eta)\|} = 0_{\mathbb R^2}$$ and consequently $$\lim_{(\xi ,\eta) \to 0_{\mathbb R^2}} \frac{F(x_0+\xi, y_0+\eta) - F(x_0, y_0) - \left[\begin{array}{cc}{\partial_{1} u\left(x_{0}, y_{0}\right)} & {-\partial_{1} v\left(x_{0}, y_{0}\right)} \\ {\partial_{1} v\left(x_{0}, y_{0}\right)} & {\partial_{1} u\left(x_{0}, y_{0}\right)}\end{array}\right] \cdot \begin{bmatrix} \xi \\ \eta \\ \end{bmatrix}}{\|(\xi ,\eta)\|} = 0_{\mathbb R^2}$$

We have the following identifications:

• $0_{\mathbb R^2} \leftrightarrow 0_{\mathbb C}$

• $(\xi ,\eta) \leftrightarrow h = \xi +i \eta$

• $F(x_0+\xi, y_0+\eta) \leftrightarrow f(z_0+h) $

• $F(x_0, y_0) \leftrightarrow f(z_0)$

• $ \left[\begin{array}{cc}{\partial_{1} u\left(x_{0}, y_{0}\right)} & {-\partial_{1} v\left(x_{0}, y_{0}\right)} \\ {\partial_{1} v\left(x_{0}, y_{0}\right)} & {\partial_{1} u\left(x_{0}, y_{0}\right)}\end{array}\right] \cdot \begin{bmatrix} \xi \\ \eta \\ \end{bmatrix} \leftrightarrow a \cdot h$

Because $\mathcal I$ preserves the topological structure, we have $$\lim _{h \rightarrow 0_{\mathbb C}} \frac{f\left(z_{0}+h\right)-f\left(z_{0}\right) - a \cdot h}{\|h\|} = 0_{\mathbb C}$$

By the definition of total derivative, we get $\partial f(z_0)h= a \cdot h$ for all $h \in \mathbb C$. Hence $f'(z_0) =a$.