Application of chain rule for a complex variable

Question

I'm looking at basic definitions in complex analysis, and I can't figure out where a factor of $1/2$ comes from below. All sources I've found either invoke it without explanation, or derive it after assuming the Cauchy-Riemann conditions.

Given $\quad f(z) = u(x,y) + i v(x,y),\quad z = x + i y $,

$\frac{\partial f}{\partial z}=\frac{\partial f}{\partial x }\frac{\partial x }{\partial z} + \frac{\partial f }{\partial y}\frac{\partial y}{\partial z}$,

Naively solving $z = x + i y$ for $x$ and $y$ and taking the partial gives:

$\frac{\partial f}{\partial z}=\frac{\partial f}{\partial x } - i\frac{\partial f }{\partial y}$.

But this is different than the correct result by a factor of $1/2$:

$\frac{\partial f}{\partial z}=\frac{1}{2}\left(\frac{\partial f}{\partial x } - i\frac{\partial f }{\partial y}\right)$.

QUESTION: Why exactly does the chain rule fail here, and how does one get the correct result without invoking Cauchy-Riemann equations first? Other illuminating remarks encouraged.

Answer 1

Let me try to clarify a confusion that was brought up in the comments, namely that $\partial x/\partial z\neq(\partial z/\partial x)^{-1}$.

Let us consider a smooth function $f:\mathbb R^n\to\mathbb R^n$ that satisfies $f(0)=0$. Near zero, $f$ can be approximated by its derivative, so that $$ f(x)\approx Df(0)x. $$ (The error term is $O(\|x\|^2)$, but let me drop it altogether.) Suppose $f^{-1}$ exists and is smooth. For it we have also $f^{-1}(y)\approx Df^{-1}(0)y$, and so $$ x=f^{-1}(f(x))\approx Df^{-1}(0) Df(0)x $$ when $\|x\|$ is small. This can only be true if $Df^{-1}(0) Df(0)$ is the identity matrix, or, in other words, $Df^{-1}(0)=Df(0)^{-1}$. Matrix inversion cannot be done element by element, so in general we have $\partial f_i/\partial x_j\neq (\partial x_j/\partial f_i)^{-1}$. The partial derivative $\partial f_i/\partial x_j$ is an element of the matrix $Df(0)$ and $\partial x_j/\partial f_i$ is an element of the matrix $Df^{-1}(0)$. The same works for functions $\mathbb C^n\to\mathbb C^n$.

In your case, we are looking at the mapping $(x,y)\mapsto(z,\bar z)=(x+iy,x-iy)$. This mapping is actually linear, so its derivative is easy to compute; it is $$ A= \begin{pmatrix} 1&i\\1&-i \end{pmatrix}. $$ The inverse of this matrix is $$ B= \frac12 \begin{pmatrix} 1&1\\-i&i \end{pmatrix}. $$ Now $\partial z/\partial x=a_{11}=1$ but $\partial x/\partial z=b_{11}=\frac12$. This is why your calculation was off by a factor of two.

A more direct way to see this would be to express $x$ in terms of $z$ and $\bar z$. From $x=\frac12(z+\bar z)$ you can see that $\partial x/\partial z=\frac12$.

I'm not sure if this is a particularly fruitful way to approach the Cauchy—Riemann operator(s), but it may give some insight. One problem is that it is not very clear how to interpret $\partial x/\partial z$ as a partial derivative.

Answer 2

Here is one approach. Convert from the two variables $x,y$ to the two variables $z, \overline{z}$. In one direction $$ z = x + i y,\qquad \overline{z} = x - i y $$ Solve to get the other direction $$ x = \frac{z+\overline{z}}{2},\qquad y=\frac{z-\overline{z}}{2i} $$ Now we need $$ \frac{\partial{x}}{\partial{z}} = \frac{1}{2},\qquad \frac{\partial y}{\partial z} = \frac{1}{2i} . $$ Put these into $\frac{\partial f}{\partial z}=\frac{\partial f}{\partial x }\frac{\partial x }{\partial z} + \frac{\partial f }{\partial y}\frac{\partial y}{\partial z}$ to get $$ \frac{\partial f}{\partial z}=\frac{1}{2}\left(\frac{\partial f}{\partial x } - i\frac{\partial f }{\partial y}\right) $$

Answer 3

Since the OP mentioned the Cauchy-Riemann equations, I am assuming that $f$ is analytic.

Now, we have both

$$f'(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x} \tag 1$$

and

$$f'(z)=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}\tag 2$$

Adding both sides of $(1)$ and $(2)$ and dividing by $2$ reveals that

$$\begin{align} f'(z)&=\frac12\left(\frac{\partial (u+iv)}{\partial x}-i\frac{\partial (u+iv)}{\partial y}\right)\\\\ &=\frac12\left(\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right) \end{align}$$

as was to be shown!

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NOTE:

We can use a method of differentials that are correct and reflect the OP's main question. To that end, we can write

$$\begin{align} df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy \end{align} \tag 3$$

Then, setting $z=x+iy$, we have $x=z-iy$ so that

$$dx=dz-idy \tag4$$

Substituting $(4)$ into $(3)$ yields

$$df=\frac{\partial f}{\partial x}dz+\left(\frac{\partial f}{\partial y}-i\frac{\partial f}{\partial x}\right)dy \tag 5$$

By setting $y=-iz+ix$, we obtain similarly

$$df=-i\frac{\partial f}{\partial y}dz+\left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right)dx\tag 6$$

For $f$ to be analytic, $(5)$ and $(6)$ imply that both (i)

$$\begin{align} \frac{\partial f}{\partial y}-i\frac{\partial f}{\partial x}&=0 \tag 7\\\\ \frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}&=0 \tag 7 \end{align}$$

from which $(7)$ gives the Cauchy-Riemann Equations, and (ii)

$$\begin{align} df&=\frac{\partial f}{\partial x}dz\\\\ &=-i\frac{\partial f}{\partial y}dz \end{align} \tag 8$$

from which $(8)$ implies that

$$\begin{align} \frac{df}{dz}&=\frac{\partial f}{\partial x}\\\\ &=-i\frac{\partial f}{\partial y}\\\\ &=\frac12\left(\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right) \end{align}$$

which recovers the expected result!