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.

  • $\begingroup$ 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
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    $\begingroup$ 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

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.

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    $\begingroup$ 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

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


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

Now, we have both

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


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

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!


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!

  • $\begingroup$ 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

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