I'm trying to understand the derivation of Wiener deconvolution given on its Wikipedia page. In the last couple steps under the derivation section, they take the derivative with respect to $G(f)$ of an equation that has both $G(f)$ and $G^\ast(f)$ in it. They simply state that $G^\ast (f)$ acts as a constant in the differentiation. However, it seems to me that if you don't treat $G(f)$ as a constant, then you shouldn't be able to treat $G^\ast (f)$ as a constant because they are directly related.

I searched around some looking for an explanation. I found this page, which seems to agree that the complex conjugate can be treated as a constant. I also found some stuff about the Cauchy-Riemann equations, which seem to be related. However, I haven't had any classes on complex analysis and don't understand the intuition behind why this can be done.

Why can the complex conjugate of a variable be treated as a constant when differentiating with respect to that variable?


The nomenclature of $\dfrac{\partial}{\partial z}$ and $\dfrac{\partial}{\partial\bar{z}}$ is confusing because it gives the impression that these are really partial derivatives with respect to two independent variables, $z$ and $\bar{z}$. However, it is clear that $z$ and $\bar{z}$ are not independent.

Differentiable Functions and Conformal Maps

A differentiable function on $\mathbb{R}$ locally looks like a linear function, that is, there is a real constant, called $f'(x)$, so that for small $h$, $$ f(x+h)=f(x)+f'(x)h+o(h)\tag{1} $$ Analogously, a differentiable function on $\mathbb{C}$ satisfies $(1)$ for some complex number $f'(x)$.

Multiplication on $\mathbb{C}$ acts as a rotation and radial scale when viewed as an action on $\mathbb{R}^2$. Thus, if $f$ is differentiable on $\mathbb{C}$, $$ f(z+h)-f(z)=f'(z)h+o(h)\tag{2} $$ That is, when $h$ is small, $h\mapsto f(z+h)-f(z)$ looks like a scaled rotation. For this reason, a differentiable function on $\mathbb{C}$ is called conformal: small features are replicated (scaled and rotated) and angles are preserved.

Complex Conjugation and Orientation Reversal

Complex conjugation, $z\mapsto\bar{z}$, is an orientation reversing isometry. Thus, when composed with a conformal map, either before or after, the composition is an orientation-reversing conformal map. Furthermore, double composition yields an orientation-preserving conformal map; for example, if $f(z)$ is conformal, then so is $\overline{f(\bar{z})}$.

As a function on $\mathbb{R}^2$, complex conjugation can be represented by the matrix $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$.

Conformal and Conjugate Conformal

The partial derivatives of a general differentiable function on $\mathbb{R}^2$ given by $x+iy\mapsto u+iv$ are usually given in a $2\times2$ Jacobian matrix: $$ \frac{\partial(u,v)}{\partial(x,y)}=\begin{bmatrix}\frac{\partial u}{\partial x}&\frac{\partial v}{\partial x}\\\frac{\partial u}{\partial y}&\frac{\partial v}{\partial y}\end{bmatrix}\tag{3} $$ The Cauchy-Riemann equations specify that $\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}$ and $\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}$, which agrees with the following basis for the orientation-preserving conformal Jacobians on $\mathbb{R}^2$: $$ \left\{\begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}0&1\\-1&0\end{bmatrix}\right\}\tag{4} $$ Note that the determinant of any linear combination of these matrices has positive determinant (thus orientation is preserved).

The following basis for the orientation-reversing conformal Jacobians on $\mathbb{R}^2$ follows by composing conjugation with $(4)$: $$ \left\{\begin{bmatrix}1&0\\0&-1\end{bmatrix},\begin{bmatrix}0&1\\1&0\end{bmatrix}\right\}\tag{5} $$ Note that the determinant of any linear combination of these matrices has negative determinant (thus orientation is reversed).

Using $(4)$ and $(5)$, we can break any Jacobian into conformal and conjugate conformal parts. Using the component-wise orthogonality that exists among the bases, we can write the conformal part as $$ \frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\begin{bmatrix}1&0\\0&1\end{bmatrix}+\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\begin{bmatrix}0&1\\-1&0\end{bmatrix}\tag{6} $$ and the conjugate conformal part as $$ \frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)\begin{bmatrix}1&0\\0&-1\end{bmatrix}+\frac{1}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)\begin{bmatrix}0&1\\1&0\end{bmatrix}\tag{7} $$ $\dfrac{\partial}{\partial z}$, $\dfrac{\partial}{\partial\bar{z}}$, and Quaternions

The definitions of $\dfrac{\partial}{\partial z}$ and $\dfrac{\partial}{\partial\bar{z}}$ say $$ \begin{align} \frac{\partial}{\partial z}(u+iv) &=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)(u+iv)\\ &=\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+\frac{i}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\tag{8} \end{align} $$ and $$ \begin{align} \frac{\partial}{\partial\bar{z}}(u+iv) &=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)(u+iv)\\ &=\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+\frac{i}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)\tag{9} \end{align} $$ The space of $2\times2$ Jacobians has $4$ dimensions, so trying to represent these $4$ dimensions with the $2$ dimensions of $\mathbb{C}$, using $\dfrac{\partial}{\partial z}$ and $\dfrac{\partial}{\partial\bar{z}}$, obscures something.

There is a common matrix representation of the complex numbers as $2\times2$ real matrices where $$ \begin{align} \mathbf{1}&\leftrightarrow\begin{bmatrix}1&0\\0&1\end{bmatrix}\tag{10}\\ \mathbf{i}&\leftrightarrow\begin{bmatrix}0&1\\-1&0\end{bmatrix}\tag{11} \end{align} $$ However, there is also a matrix representation of the quaternions as $2\times2$ complex matrices where, in addition to $(10)$ and $(11)$, $$ \begin{align} \mathbf{j}&\leftrightarrow\begin{bmatrix}i&0\\0&-i\end{bmatrix}\tag{12}\\ \mathbf{k}&\leftrightarrow\begin{bmatrix}0&-i\\-i&0\end{bmatrix}\tag{13} \end{align} $$ Embed $(8)$ and $(9)$ in the quaternions to get $$ \left(\frac{\partial}{\partial z}(u+iv)\right)\mathbf{1} =\frac{\mathbf{1}}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+\frac{\mathbf{i}}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\tag{14} $$ and $$ \left(\frac{\partial}{\partial\bar{z}}(u+iv)\right)\mathbf{j} =\frac{\mathbf{j}}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+\frac{\mathbf{k}}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)\tag{15} $$ Finally, substituting $(10)$-$(13)$ into $(14)$ and $(15)$, it becomes apparent, upon comparison with $(6)$ and $(7)$, that $\left(\dfrac{\partial}{\partial z}(u+iv)\right)\mathbf{1}$ represents the conformal part of the Jacobian and $\left(\dfrac{\partial}{\partial\bar{z}}(u+iv)\right)\mathbf{j}$ represents the conjugate conformal part.


For a general $f:\mathbb{C}\mapsto\mathbb{C}$, $\dfrac{\partial}{\partial z}f$ can be mapped to the conformal part of the $2\times2$ Jacobian, $\dfrac{\partial f}{\partial z}=\dfrac{\partial(u,v)}{\partial(x,y)}$, and $\dfrac{\partial}{\partial\bar{z}}f$ can be mapped to the conjugate conformal part. It is merely convenience of notation that we write $\dfrac{\partial}{\partial z}$ and $\dfrac{\partial}{\partial\bar{z}}$ because $\dfrac{\partial}{\partial z}f\;\mathrm{d}z+\dfrac{\partial}{\partial\bar{z}}f\;\mathrm{d}\bar{z}=\mathrm{d}f$. However, they are not true partial derivatives, but $2$ pieces of a $2\times2$ Jacobian composed of $4$ partial derivatives.

So, to answer the question asked, $z\mapsto\bar{z}$ is conjugate conformal, so $\frac{\partial}{\partial z}\bar{z}=0$; therefore, $\bar{z}$ acts like a constant under $\frac{\partial}{\partial z}$.

  • $\begingroup$ @robjohn The quaternion matrix representation on the wikipedia page you linked is not the same you used, if I am not mistaken. Also, shouldn't ij=k ? For your representation of quaternions, I seem to get ij = -k . I think the matrices for j and k should be switched. $\endgroup$ – Konstantin Schubert Nov 12 '13 at 19:54
  • $\begingroup$ @Konstantin: I changed the sign of $\mathbf{k}$. That should fix things, I think. $\endgroup$ – robjohn Nov 12 '13 at 21:44

$\displaystyle{\frac{\partial}{\partial z}}$ and $\displaystyle{\frac{\partial}{\partial \overline z}}$ are defined such that $\displaystyle{\frac{\partial}{\partial z} z=1}$, $\displaystyle{\frac{\partial}{\partial z}\overline{z}=0}$, $\displaystyle{\frac{\partial}{\partial \overline z} z=0}$, and $\displaystyle{\frac{\partial}{\partial \overline z}\overline z=1}$. This shows why $\overline{z}$ can be treated as constant when differentiating with respect to $z$, in that differentiating something like $z\overline z + 3\overline{z}^3$ with respect to $z$ is very similar to differentiating $xy+3y^3$ with respect to $x$.

In terms of real coordinates, $z=x+iy$, $\overline z=x-iy$, $\displaystyle{\frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}- i\frac{\partial}{\partial y}\right)}$ and $\displaystyle{\frac{\partial}{\partial \overline z}=\frac{1}{2}\left(\frac{\partial}{\partial x}+ i\frac{\partial}{\partial y}\right)}$.

The Cauchy-Riemann equations are related, because they are equivalent to $\displaystyle{\frac{\partial}{\partial \overline z}f=0}$.


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