1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Because you cannot solve $z=x+i y$ for x and for y, you need another equation: $\bar z = x - i y$, for the system to have a solution! $\endgroup$ – krvolok Jul 7 '15 at 16:22
  • 1
    $\begingroup$ What do you mean by $\partial x/\partial z$? $\endgroup$ – Joonas Ilmavirta Jul 7 '15 at 16:24
  • $\begingroup$ @Joonas: $x = z - i y, \ \frac{\partial x}{\partial z} = 1$. I'm pretty sure there's an error here, I'm looking for a formal explanation of why this isn't correct. $\endgroup$ – anon01 Jul 7 '15 at 16:30
  • $\begingroup$ In general $(\partial z/\partial x)^{-1}\neq\partial x/\partial z$. To find the derivative of the inverse, you should be inverting a $2\times 2$ matrix of derivatives, not a single partial derivative. This actually leads to an error of a factor of two in this case. $\endgroup$ – Joonas Ilmavirta Jul 7 '15 at 16:32
  • $\begingroup$ Ok, this sounds like a path to clarity. Can you put this in an answer? $\endgroup$ – anon01 Jul 7 '15 at 16:37
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ In hindsight it's pretty obvious that two independent variables ($x,y$) should map to two independent complex variables ($z,\bar z$). Strange that I've never seen this covered in a complex analysis course. Thanks! $\endgroup$ – anon01 Jul 8 '15 at 1:09
5
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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!


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!

$\endgroup$
  • $\begingroup$ Yes, thanks - I've seen this via Arfken & Weber, my question is really where I've misstepped to get an incorrect answer. $\endgroup$ – anon01 Jul 7 '15 at 16:35
  • $\begingroup$ You're welcome. Should I delete this then? $\endgroup$ – Mark Viola Jul 7 '15 at 16:42
  • $\begingroup$ Please leave this - it is a useful way to arrive at the correct answer - it simply doesn't give an indication where I went wrong. $\endgroup$ – anon01 Jul 7 '15 at 18:27
  • $\begingroup$ OK. I shall. Have you found the source of error? $\endgroup$ – Mark Viola Jul 7 '15 at 18:29
  • $\begingroup$ Not exactly. All answers provided indicate that I must consider pairs u,v or z, $\bar z$ when computing the differential, and Joonas' answer goes a step further by considering the mapping between $x,y$ and $z,\bar z$. However, if I write down $x = z - i y$ and take the partial with respect to $z$, what exactly have I violated? $x,y$ are independent. Without a priori knowledge, how would I know that the other relevant variable for mapping is $\bar z$ (@ Joonas) or to rewrite this expression in terms of $\bar z$ (@ GEdgar)? Can one map to any set of linearly independent variables? $\endgroup$ – anon01 Jul 7 '15 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.