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

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.

Conclusion

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 $\frac{\partial}{\partial z}$.

• @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. – Konstantin Schubert Nov 12 '13 at 19:54
• @Konstantin: I changed the sign of $\mathbf{k}$. That should fix things, I think. – 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 $\displaystyle{\frac{\partial}{\partial \overline z}f=0}$.