closure properties of Cauchy-Riemann equations

The set of solutions of the Cauchy-Riemann equations is closed under a surprisingly large number of operations. For convenience, let us introduce the notational conventions that $f$ and $g$ are complex functions with $f(x+iy)=u(x,y)+iv(x,y)$ and $g(x+iy)=p(x,y)+iq(x,y)$. Let $D$ and $D^{\prime }$ denote open subsets of the complex plane.

Theorem 1.

If $f\mathrm{:}D\mathrm{\to }\mathrm{C}$ and $g\mathrm{:}D\mathrm{\to }\mathrm{C}$ satisfy the Cauchy-Riemann equations, so does $f\mathrm{+}g$. Furthermore, if $z\mathrm{\in }\mathrm{C}$, then $z\mathit{}f$ satisfies the Cauchy-Riemann equations.

Proof.

This is an immediate consequence of the linearity of derivatives. ∎

Theorem 2.

If $f\mathrm{:}D\mathrm{\to }\mathrm{C}$ and $g\mathrm{:}D\mathrm{\to }\mathrm{C}$ satisfy the Cauchy-Riemann equations, so does $f\mathrm{\cdot }g$.

Proof.

Letting $h$ and $k$ denote the real and imaginary parts of $f\cdot g$ respectively, we have

${\displaystyle \frac{\partial h}{\partial x}}-{\displaystyle \frac{\partial k}{\partial y}}$ $={\displaystyle \frac{\partial }{\partial x}}\left(up-vq\right)-{\displaystyle \frac{\partial }{\partial y}}\left(uq+vp\right)$

$=u{\displaystyle \frac{\partial p}{\partial x}}+p{\displaystyle \frac{\partial u}{\partial x}}-v{\displaystyle \frac{\partial q}{\partial x}}-q{\displaystyle \frac{\partial v}{\partial x}}-u{\displaystyle \frac{\partial q}{\partial y}}-q{\displaystyle \frac{\partial u}{\partial y}}-v{\displaystyle \frac{\partial p}{\partial y}}-p{\displaystyle \frac{\partial v}{\partial y}}$

$=u\left({\displaystyle \frac{\partial p}{\partial x}}-{\displaystyle \frac{\partial q}{\partial y}}\right)-v\left({\displaystyle \frac{\partial p}{\partial y}}+{\displaystyle \frac{\partial q}{\partial x}}\right)+p\left({\displaystyle \frac{\partial u}{\partial x}}-{\displaystyle \frac{\partial v}{\partial y}}\right)-q\left({\displaystyle \frac{\partial u}{\partial y}}+{\displaystyle \frac{\partial v}{\partial x}}\right)=0$

and

${\displaystyle \frac{\partial h}{\partial y}}+{\displaystyle \frac{\partial k}{\partial x}}$ $={\displaystyle \frac{\partial }{\partial y}}\left(up-vq\right)+{\displaystyle \frac{\partial }{\partial x}}\left(uq+vp\right)$

$=u{\displaystyle \frac{\partial p}{\partial y}}+p{\displaystyle \frac{\partial u}{\partial y}}-v{\displaystyle \frac{\partial q}{\partial y}}-q{\displaystyle \frac{\partial v}{\partial y}}+u{\displaystyle \frac{\partial q}{\partial x}}+q{\displaystyle \frac{\partial u}{\partial x}}+v{\displaystyle \frac{\partial p}{\partial x}}+p{\displaystyle \frac{\partial v}{\partial x}}$

$=u\left({\displaystyle \frac{\partial p}{\partial y}}+{\displaystyle \frac{\partial q}{\partial x}}\right)+v\left({\displaystyle \frac{\partial p}{\partial x}}-{\displaystyle \frac{\partial q}{\partial y}}\right)+p\left({\displaystyle \frac{\partial u}{\partial y}}+{\displaystyle \frac{\partial v}{\partial x}}\right)+q\left({\displaystyle \frac{\partial u}{\partial x}}-{\displaystyle \frac{\partial v}{\partial y}}\right)=0.$

∎

Theorem 3.

If $f\mathrm{:}D\mathrm{\to }{D}^{\mathrm{\prime }}$ and $g\mathrm{:}{D}^{\mathrm{\prime }}\mathrm{\to }\mathrm{C}$ satisfy the Cauchy-Riemann equations, so does $f\mathrm{\circ }g$.

Proof.

Letting $h$ and $k$ denote the real and imaginary parts of $f\circ g$ respectively, we have

${\displaystyle \frac{\partial h}{\partial x}}-{\displaystyle \frac{\partial k}{\partial y}}$ $={\displaystyle \frac{\partial }{\partial x}}u(p(x,y),q(x,y))-{\displaystyle \frac{\partial }{\partial y}}v(p(x,y),q(x,y))$

$={\displaystyle \frac{\partial u}{\partial p}}{\displaystyle \frac{\partial p}{\partial x}}+{\displaystyle \frac{\partial u}{\partial q}}{\displaystyle \frac{\partial q}{\partial x}}-{\displaystyle \frac{\partial v}{\partial p}}{\displaystyle \frac{\partial p}{\partial y}}-{\displaystyle \frac{\partial v}{\partial q}}{\displaystyle \frac{\partial q}{\partial y}}$

$={\displaystyle \frac{\partial u}{\partial p}}\left({\displaystyle \frac{\partial p}{\partial x}}-{\displaystyle \frac{\partial q}{\partial y}}\right)+{\displaystyle \frac{\partial q}{\partial y}}\left({\displaystyle \frac{\partial u}{\partial p}}-{\displaystyle \frac{\partial v}{\partial q}}\right)+{\displaystyle \frac{\partial u}{\partial q}}\left({\displaystyle \frac{\partial p}{\partial y}}+{\displaystyle \frac{\partial q}{\partial x}}\right)-{\displaystyle \frac{\partial p}{\partial y}}\left({\displaystyle \frac{\partial u}{\partial q}}+{\displaystyle \frac{\partial v}{\partial p}}\right)=0$

and

${\displaystyle \frac{\partial h}{\partial y}}+{\displaystyle \frac{\partial k}{\partial x}}$ $={\displaystyle \frac{\partial }{\partial y}}u(p(x,y),q(x,y))+{\displaystyle \frac{\partial }{\partial x}}v(p(x,y),q(x,y))$

$={\displaystyle \frac{\partial u}{\partial p}}{\displaystyle \frac{\partial p}{\partial y}}+{\displaystyle \frac{\partial u}{\partial q}}{\displaystyle \frac{\partial q}{\partial y}}+{\displaystyle \frac{\partial v}{\partial p}}{\displaystyle \frac{\partial p}{\partial x}}+{\displaystyle \frac{\partial v}{\partial q}}{\displaystyle \frac{\partial q}{\partial x}}$

$={\displaystyle \frac{\partial u}{\partial p}}\left({\displaystyle \frac{\partial p}{\partial y}}+{\displaystyle \frac{\partial q}{\partial x}}\right)-{\displaystyle \frac{\partial q}{\partial x}}\left({\displaystyle \frac{\partial u}{\partial p}}-{\displaystyle \frac{\partial v}{\partial q}}\right)-{\displaystyle \frac{\partial u}{\partial q}}\left({\displaystyle \frac{\partial p}{\partial x}}-{\displaystyle \frac{\partial q}{\partial y}}\right)+{\displaystyle \frac{\partial p}{\partial x}}\left({\displaystyle \frac{\partial u}{\partial q}}+{\displaystyle \frac{\partial v}{\partial p}}\right)=0$

∎

Theorem 4.

If $f\mathrm{:}D\mathrm{\to }\mathrm{C}$ satisfies the Cauchy-Riemann equations, and has non-vanishing Jacobian, then $f^{\mathrm{-}\mathrm{1}}$ also satisfies the Cauchy-Riemann equations.

Proof.

Let us denote the real and imaginary parts of $f^{-1}$ as $h$ and $k$, respectively. Then, by definition of inverse function, we have

$u(h(x,y),k(x,y))$ $=x$

$v(h(x,y),k(x,y))$ $=y.$

Taking derivatives,

${\displaystyle \frac{\partial u}{\partial h}}{\displaystyle \frac{\partial h}{\partial x}}+{\displaystyle \frac{\partial u}{\partial k}}{\displaystyle \frac{\partial k}{\partial x}}$ $=1$

${\displaystyle \frac{\partial u}{\partial h}}{\displaystyle \frac{\partial h}{\partial y}}+{\displaystyle \frac{\partial u}{\partial k}}{\displaystyle \frac{\partial k}{\partial y}}$ $=0$

${\displaystyle \frac{\partial v}{\partial h}}{\displaystyle \frac{\partial h}{\partial x}}+{\displaystyle \frac{\partial v}{\partial k}}{\displaystyle \frac{\partial k}{\partial x}}$ $=0$

${\displaystyle \frac{\partial v}{\partial h}}{\displaystyle \frac{\partial h}{\partial y}}+{\displaystyle \frac{\partial v}{\partial k}}{\displaystyle \frac{\partial k}{\partial y}}$ $=1$

By the Cauchy-Riemann equations, $\partial u/\partial h=\partial v/\partial k$ and $\partial u/\partial k=-\partial v/\partial h$. Using these relations to re-express the derivatives of $u$ as derivatives of $v$, then subtracting the fourth equation form the first equation and adding the second and third equations, we obtain

${\displaystyle \frac{\partial u}{\partial h}}\left({\displaystyle \frac{\partial h}{\partial x}}-{\displaystyle \frac{\partial k}{\partial y}}\right)+{\displaystyle \frac{\partial u}{\partial k}}\left({\displaystyle \frac{\partial h}{\partial y}}+{\displaystyle \frac{\partial k}{\partial x}}\right)$ $=0$

${\displaystyle \frac{\partial u}{\partial h}}\left({\displaystyle \frac{\partial h}{\partial y}}+{\displaystyle \frac{\partial k}{\partial x}}\right)-{\displaystyle \frac{\partial u}{\partial k}}\left({\displaystyle \frac{\partial h}{\partial x}}-{\displaystyle \frac{\partial k}{\partial y}}\right)$ $=0.$

With a little algebraic manipulation, we may conclude

$\left({\left({\displaystyle \frac{\partial u}{\partial h}}\right)}^2+{\left({\displaystyle \frac{\partial u}{\partial k}}\right)}^2\right)\left({\displaystyle \frac{\partial h}{\partial y}}+{\displaystyle \frac{\partial k}{\partial x}}\right)$ $=0$

$\left({\left({\displaystyle \frac{\partial u}{\partial h}}\right)}^2+{\left({\displaystyle \frac{\partial u}{\partial k}}\right)}^2\right)\left({\displaystyle \frac{\partial h}{\partial x}}-{\displaystyle \frac{\partial k}{\partial y}}\right)$ $=0.$

Note that, by the Cauchy-Riemann equations, the Jacobian of $f$ equals the common prefactor of these equations:

$$\frac{\partial (u,v)}{\partial (h,k)}=\frac{\partial u}{\partial h}\frac{\partial v}{\partial k}-\frac{\partial u}{\partial k}\frac{\partial v}{\partial h}={\left(\frac{\partial u}{\partial h}\right)}^2+{\left(\frac{\partial u}{\partial k}\right)}^2$$

Hence, by assumptions, this quantity differs from zero and we may cancel it to obtain the Cauchy-Riemann equations for $f^{-1}$. ∎

Title	closure properties of Cauchy-Riemann equations
Canonical name	ClosurePropertiesOfCauchyRiemannEquations
Date of creation	2013-03-22 17:44:20
Last modified on	2013-03-22 17:44:20
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	14
Author	rspuzio (6075)
Entry type	Theorem
Classification	msc 30E99
Related topic	TangentialCauchyRiemannComplexOfCinftySmoothForms
Related topic	ACRcomplex